算法题解之图论与搜索

Additive Number

加法数

思路：一开始以为要用DP来做，但是这道题除非能保证每个相加数的划分是唯一的（事实上不是唯一的），否则只能用搜索。      

public class Solution {
    public boolean isAdditiveNumber(String num) {
        for (int firstEnd = 0; firstEnd <= num.length() - 3; firstEnd++) {
            if (num.charAt(0) == '0' && firstEnd > 0) {
                break;
            }
            for (int secondEnd = firstEnd + 1; secondEnd <= num.length() - 2; secondEnd++) {
                if (num.charAt(firstEnd + 1) == '0' && secondEnd > firstEnd + 1) {
                    break;
                }
                long first = Long.parseLong(num.substring(0, firstEnd + 1));
                long second = Long.parseLong(num.substring(firstEnd + 1, secondEnd + 1));
                if (helper(num, secondEnd + 1, first + second, second)) {
                    return true;
                }
            }
        }
        return false;
    }
    
    public boolean helper(String num, int start, long expect_num, long pre_num) {
        if (start > num.length() - 1) {
            return true;
        }
        if (num.charAt(start) == '0' && expect_num != 0) {
            return false;
        }
        int end = start + String.valueOf(expect_num).length() - 1;
        if (end + 1 > num.length()) {
            return false;
        }
        long cur = Long.parseLong(num.substring(start, end + 1));
        if (cur == expect_num) {
            long next_expect = pre_num + cur;
            return helper(num, end + 1, next_expect, cur);
        }
        return false;
    }
}

 

 

Clone Graph

克隆图

思路：类似于copy random list，先克隆每个节点的值，建立新老节点的hash映射，再根据映射关系克隆邻接表。

/**
 * Definition for undirected graph.
 * class UndirectedGraphNode {
 *     int label;
 *     ArrayList<UndirectedGraphNode> neighbors;
 *     UndirectedGraphNode(int x) { label = x; neighbors = new ArrayList<UndirectedGraphNode>(); }
 * };
 */
public class Solution {
    /**
     * @param node: A undirected graph node
     * @return: A undirected graph node
     */
    public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
        // write your code here
        if (node == null) {
            return null;
        }
        Map<UndirectedGraphNode, UndirectedGraphNode> old2new 
                = new HashMap<UndirectedGraphNode, UndirectedGraphNode>();
        Queue<UndirectedGraphNode> q = new LinkedList<UndirectedGraphNode>();
        
        q.offer(node);
        UndirectedGraphNode copy = new UndirectedGraphNode(node.label);
        copy.neighbors = new ArrayList<UndirectedGraphNode>(node.neighbors);
        old2new.put(node, copy);
        while (!q.isEmpty()) {
            int size = q.size();
            for (int i = 1; i <= size; i++) {
                UndirectedGraphNode cur = q.poll();
                for (UndirectedGraphNode next : cur.neighbors) {
                    if (!old2new.containsKey(next)) {
                        q.offer(next);
                        UndirectedGraphNode copyNode = new UndirectedGraphNode(next.label);
                        copyNode.neighbors = new ArrayList<UndirectedGraphNode>(next.neighbors);
                        old2new.put(next, copyNode);
                    }
                }
            }
        }
        
        for (Map.Entry<UndirectedGraphNode, UndirectedGraphNode> entry : old2new.entrySet()) {
            UndirectedGraphNode copy2 = entry.getValue();
            ArrayList<UndirectedGraphNode> neighbors = copy2.neighbors; 
            for (int i = 0; i < neighbors.size(); i++) {
                neighbors.set(i, old2new.get(neighbors.get(i)));
            }
        }
        
        return old2new.get(node);
    }
}

 

 

Course Schedule

课程计划

思路：这道题的本质其实是判断一张图是否存在拓扑排序。拓扑排序表示的是修课的顺序。根据边集合构造图，然后就是拓扑排序那一套了。

public class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        Map<Integer, Value> map = new HashMap<Integer, Value>();
        for (int i = 0; i < prerequisites.length; i++) {
            int to = prerequisites[i][0];
            int from = prerequisites[i][1];
            
            if (!map.containsKey(from)) {
                map.put(from, new Value());
            }
            if (!map.containsKey(to)) {
                map.put(to, new Value());
            }
            map.get(from).neighbors.add(to);
            map.get(to).inEdges += 1; 
        }
        
        
        Queue<Integer> q = new LinkedList<Integer>();
        for (Integer course : map.keySet()) {
            if (map.get(course).inEdges == 0) {
                q.offer(course);
            }
        }
        int notFinished = map.size();
        while (!q.isEmpty()) {
            int finish = q.poll();
            notFinished--;
            for (Integer course : map.get(finish).neighbors) {
                if (--map.get(course).inEdges == 0) {
                    q.offer(course);   
                }
            }
        }
        if (notFinished == 0) {
            return true;
        } else {
            return false;
        }
    }
    
}

class Value {
    int inEdges;
    List<Integer> neighbors;
    Value() {
        neighbors = new ArrayList<Integer>();
    }
}

 

 

Course Schedule II

课程计划II

思路：本质就是求拓扑排序。

public class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        Map<Integer, Value> map = new HashMap<Integer, Value>();
        for (int i = 0; i < prerequisites.length; i++) {
            int to = prerequisites[i][0];
            int from = prerequisites[i][1];
            
            if (!map.containsKey(from)) {
                map.put(from, new Value());
            }
            if (!map.containsKey(to)) {
                map.put(to, new Value());
            }
            map.get(from).neighbors.add(to);
            map.get(to).inEdges += 1; 
        }
        
        Queue<Integer> q = new LinkedList<Integer>();
        for (Integer course : map.keySet()) {
            if (map.get(course).inEdges == 0) {
                q.offer(course);
            }
        }
        
        int[] res = new int[numCourses];
        int cur = 0;
        while (!q.isEmpty()) {
            int finish = q.poll();
            res[cur++] = finish;
            for (Integer course : map.get(finish).neighbors) {
                if (--map.get(course).inEdges == 0) {
                    q.offer(course);   
                }
            }
        }
        if (cur != map.size()) {
            return new int[0];
        }
        
        for (int i = 0; i < numCourses; i++) {
            if (!map.containsKey(i)) {
                res[cur++] = i;
            }
        }

        return res;
    }
    
}

class Value {
    int inEdges;
    List<Integer> neighbors;
    Value() {
        neighbors = new ArrayList<Integer>();
    }
}

 

 

Combination Sum

组合sum

思路：常规DFS题，先排序，再搜索。注意代码中如何去重。

public class Solution {
    /**
     * @param candidates: A list of integers
     * @param target:An integer
     * @return: A list of lists of integers
     */
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        // write your code here
        Arrays.sort(candidates);
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        helper(res, new ArrayList<Integer>(), candidates, 0, 0, target);
        return res;
    }
    
    public void helper(List<List<Integer>> res, List<Integer> cur, 
        int[] candidates, int start, int sum, int target) {
            
        if (sum == target) {
            List<Integer> tmp = new ArrayList<Integer>(cur);
            res.add(tmp);
            return;
        }
        
        for (int i = start; i < candidates.length;) {
            if (sum + candidates[i] > target) {
                break;
            }
            cur.add(candidates[i]);
            helper(res, cur, candidates, i, sum + candidates[i], target);
            cur.remove(cur.size() - 1);
            
            if(i == candidates.length - 1) {
                break;
            }
            while (i < candidates.length - 1 && candidates[i++] == candidates[i]) {}
            if (i == candidates.length - 1 && candidates[i] == candidates[i-1]) {
                i++;
            }
        }
    }
}

 

 

Find the Connected Component in the Undirected Graph

找无向图的连通块

思路：以图中任意一个节点为起点做搜索，搜索完就找到一个连通块，并把搜索过的节点加入set。再对图中其他节点，如果搜索过就pass，没搜索过就搜索。

/**
 * Definition for Undirected graph.
 * class UndirectedGraphNode {
 *     int label;
 *     ArrayList<UndirectedGraphNode> neighbors;
 *     UndirectedGraphNode(int x) { label = x; neighbors = new ArrayList<UndirectedGraphNode>(); }
 * };
 */
public class Solution {
    /**
     * @param nodes a array of Undirected graph node
     * @return a connected set of a Undirected graph
     */
    public List<List<Integer>> connectedSet(ArrayList<UndirectedGraphNode> nodes) {
        // Write your code here
        Set<UndirectedGraphNode> searchedNodes = new HashSet<UndirectedGraphNode>();
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        for (UndirectedGraphNode node : nodes) {
            if (searchedNodes.contains(node)) {
                continue;                
            }
            List<Integer> cur_res = new ArrayList<Integer>();
            Queue<UndirectedGraphNode> q = new LinkedList<UndirectedGraphNode>();
            Set<UndirectedGraphNode> set = new HashSet<UndirectedGraphNode>();
            q.offer(node);
            set.add(node);
            searchedNodes.add(node);
            while (!q.isEmpty()) {
                int size = q.size();
                for (int i = 1; i <= size; i++) {
                    UndirectedGraphNode cur = q.poll();
                    cur_res.add(cur.label);
                    for (UndirectedGraphNode neighbor : cur.neighbors) {
                        if (!set.contains(neighbor)) {
                            q.offer(neighbor);
                            set.add(neighbor);
                            searchedNodes.add(neighbor);
                        }
                    }
                }
            }
            Collections.sort(cur_res);
            res.add(cur_res);
        }
        return res;
    }
}

 

 

k Sum II

k数和II

思路：常规的搜索题。注意求方案个数的K sum是用动态规划。

public class Solution {
    /**
     * @param A: an integer array.
     * @param k: a positive integer (k <= length(A))
     * @param target: a integer
     * @return a list of lists of integer 
     */ 
    public ArrayList<ArrayList<Integer>> kSumII(int[] A, int k, int target) {
        // write your code here
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        helper(res, new ArrayList<Integer>(), 0, target, k, 0, A);
        return res;
    }
    
    public void helper(ArrayList<ArrayList<Integer>> res, ArrayList<Integer> cur, 
        int start, int target, int k, int sum, int[] A) {
        
        if (cur.size() == k) {
            if (sum == target) {
                res.add(new ArrayList<Integer>(cur));
            }
            return;
        }
        
        for (int i = start; i < A.length; i++) {
            cur.add(A[i]);
            helper(res, cur, i + 1, target, k, sum + A[i], A);
            cur.remove(cur.size() - 1);
        }
    }
}

 

 

Minimum Height Trees 

最小高度树

思路：不断去掉外面一层叶子节点（即度为1的点），最后剩下的一个或两个点就是root。原理暂时还不懂。。

public class Solution {
    public List<Integer> findMinHeightTrees(int n, int[][] edges) {
        List<Integer> res = new ArrayList<Integer>();
        if (n == 1) {
            res.add(0);
            return res;
        }
        if (n == 2) {
            res.add(0);
            res.add(1);
            return res;
        }
        
        List<Set<Integer>> map = new ArrayList<Set<Integer>>();
        for (int i = 0; i < edges.length; i++) {
            int n1 = edges[i][0];
            int n2 = edges[i][1];
            if (map.get(n1) == null) {
                map.add(n1, new HashSet<Integer>());
            }
            if (map.get(n2) == null) {
                map.put(n2, new HashSet<Integer>());
            }
            map.get(n1).add(n2);
            map.get(n2).add(n1);
        }
        
        Queue<Integer> q = new LinkedList<Integer>();
        for (Integer node : map.keySet()) {
            if (map.get(node).size() == 1) {
                q.offer(node);
            }
        }
        
        while (!q.isEmpty()) {
            int size = q.size();
            for (int i = 1; i <= size; i++) {
                int cur = q.poll();
                int neighbor = map.get(cur).iterator().next();
                map.get(neighbor).remove(cur);
                if (map.get(neighbor).size() == 1) {
                    q.offer(neighbor);
                }
                map.remove(cur);
            }
            if (map.size() <= 2) {
                break;
            }
        }
        
        for (Integer node : map.keySet()) {
            res.add(node);
        }
        return res;
    }
}

 

 

Number of Islands

小岛数量

思路：遍历每个元素，如果是岛屿并且没有被搜过就BFS这个岛，同时岛屿数加一。 

public class Solution {
    public int numIslands(char[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0) {
            return 0;
        }
        int res = 0;
        boolean[][] searched = new boolean[grid.length][grid[0].length];
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == '1' && !searched[i][j]) {
                    res++;
                    searchAndMark(grid, i, j, searched);
                }
            }
        }
        return res;
    }
    
    public void searchAndMark(char[][] grid, int cur_x, int cur_y, boolean[][] searched) {
        
        if (cur_x >= 0 && cur_x < grid.length && cur_y >= 0 && cur_y < grid[0].length 
                && !searched[cur_x][cur_y] && grid[cur_x][cur_y] == '1') {
            searched[cur_x][cur_y] = true;
            searchAndMark(grid, cur_x - 1, cur_y, searched);
            searchAndMark(grid, cur_x, cur_y - 1, searched);
            searchAndMark(grid, cur_x + 1, cur_y, searched);
            searchAndMark(grid, cur_x, cur_y + 1, searched);
        }
    }
}

 

 

N-Queens

N皇后

思路：常规DFS。使用三个boolean数组分别记录每一列，每一条主对角线（i-j相等），每一条副对角线（i+j相等）是否能放。再用一个字符数组board表示当前的棋牌放置。

class Solution {
    /**
     * Get all distinct N-Queen solutions
     * @param n: The number of queens
     * @return: All distinct solutions
     * For example, A string '...Q' shows a queen on forth position
     */
    ArrayList<ArrayList<String>> solveNQueens(int n) {
        // write your code here
        char[][] board = new char[n][n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                board[i][j] = '.';
            }
        }
        
        boolean[] col_used = new boolean[n];
        boolean[] pos_diag_used = new boolean[2 * n]; 
        boolean[] neg_diag_used = new boolean[2 * n];
        ArrayList<ArrayList<String>> res = new ArrayList<ArrayList<String>>();
        dfs(res, board, col_used, pos_diag_used, neg_diag_used, 0);
        return res;
    }
    
    public void dfs(ArrayList<ArrayList<String>> res, char[][] board, 
        boolean[] col_used, boolean[] pos_diag_used, boolean[] neg_diag_used,
        int cur_row) {
        int n = board.length;
        if (cur_row == n) {
            ArrayList<String> curSolution = buildRes(board);
            res.add(curSolution);
            return;
        }
        
        for (int j = 0; j < n; j++) {
            int pos_diag_index = (cur_row - j >= 0) ? cur_row - j  : 
                n - 1 + j - cur_row;
            if (col_used[j] == false && pos_diag_used[pos_diag_index] == false &&
                neg_diag_used[cur_row + j] == false) {
                
                col_used[j] = true; 
                pos_diag_used[pos_diag_index] = true;
                neg_diag_used[cur_row + j] = true;
                board[cur_row][j] = 'Q';
                dfs(res, board, col_used, pos_diag_used, neg_diag_used, cur_row + 1);
                col_used[j] = false; 
                pos_diag_used[pos_diag_index] = false;
                neg_diag_used[cur_row + j] = false;
                board[cur_row][j] = '.';
            }
        }
    }
    
    public ArrayList<String> buildRes(char[][] board) {
        ArrayList<String> solution = new ArrayList<String>();
        for (int i = 0; i < board.length; i++) {
            solution.add(String.valueOf(board[i]));
        }
        return solution;
    }
};

 

 

N-Queens II

N皇后II

思路：常规DFS搜索题。使用三个boolean数组分别记录每一列，每一条主对角线（i-j相等），每一条副对角线（i+j相等）是否能放。

class Solution {
    /**
     * Calculate the total number of distinct N-Queen solutions.
     * @param n: The number of queens.
     * @return: The total number of distinct solutions.
     */
    int solutionCount = 0;
    public int totalNQueens(int n) {
        //write your code here
        boolean[] col_Used = new boolean[n];
        boolean[] diag_pos_used = new boolean[2 * n]; 
        boolean[] diag_neg_used = new boolean[2 * n]; 
        dfs(col_Used, diag_pos_used, diag_neg_used, 0, n);
        return solutionCount;
    }
    
    public void dfs(boolean[] col_Used, boolean[] diag_pos_used, 
        boolean[] diag_neg_used, int cur_row, int n) {
        
        if (cur_row == n) {
            solutionCount++;
            return;
        }
        for (int i = 0; i < n; i++) {
            int tmp = (cur_row - i) >= 0 ? cur_row - i : n-1+(i-cur_row);  
            if (col_Used[i] == false && diag_pos_used[tmp] == false &&
                diag_neg_used[i+cur_row] == false) {
                col_Used[i] = true;
                diag_pos_used[tmp] = true;
                diag_neg_used[i+cur_row] = true;
                dfs(col_Used, diag_pos_used, diag_neg_used, cur_row + 1, n);
                col_Used[i] = false;
                diag_pos_used[tmp] = false;
                diag_neg_used[i+cur_row] = false;
            }
        }
    }  
}

 

 

Palindrome Partitioning

分割回文串

思路：时间复杂度是O(2^n)，即每个切口切或不切，有2^n-1种切法。先用动态规划求出一个二维数组表示每个字串是否是回文串，再进行DFS。

public class Solution {
    /**
     * @param s: A string
     * @return: A list of lists of string
     */
    public List<List<String>> partition(String s) {
        // write your code here
        int[][] dp = new int[s.length()][s.length()];
        for (int i = 0; i <= s.length() - 1; i++) {
            dp[i][i] = 1;
        }
        for (int i = 0; i <= s.length() - 2; i++) {
            if (s.charAt(i) == s.charAt(i + 1)) {
                dp[i][i+1] = 1;
            }
        }
        for (int len = 3; len <= s.length(); len++) {
            for (int i = 0; i <= s.length() - len; i++) {
                if (s.charAt(i) == s.charAt(i+len-1) && dp[i+1][i+len-2] == 1) {
                    dp[i][i+len-1] = 1;
                }
            }
        }
        
        List<List<String>> res = new ArrayList<List<String>>();
        helper(res, new ArrayList<String>(), s, 0, dp);
        return res;
    }
    
    public void helper(List<List<String>> res, List<String> cur, String s, 
                        int start, int[][] dp) {
        if (start > s.length() - 1) {
            List<String> tmp = new ArrayList<String>(cur);
            res.add(tmp);
            return;
        }
        
        for (int end = start; end <= s.length() - 1; end++) {
            if (dp[start][end] == 1) {
                cur.add(s.substring(start, end + 1));
                helper(res, cur, s, end + 1, dp);
                cur.remove(cur.size() - 1);
            }
        }
    }
}

 

 

Permutations II

带重复元素的排列

思路：常规DFS。记录一个used数组。如何避免重复： 比如，给出一个排好序的数组，[1,2,2]，那么第一个2和第二2如果在结果中互换位置， 我们也认为是同一种方案，所以我们强制要求相同的数字，原来排在前面的，在结果当中也应该排在前面，这样就保证了唯一性。所以当前面的2还没有使用的时 候，就不应该让后面的2使用。

class Solution {
    /**
     * @param nums: A list of integers.
     * @return: A list of unique permutations.
     */
    public List<List<Integer>> permuteUnique(int[] nums) {
        // Write your code here
        Arrays.sort(nums);
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        helper(res, new ArrayList<Integer>(), new boolean[nums.length], nums);
        return res;
    }
    
    public void helper(List<List<Integer>> res, List<Integer> cur, 
        boolean[] used, int[] nums) {
        
        if (cur.size() == nums.length) {
            List<Integer> tmp = new ArrayList<Integer>(cur);
            res.add(tmp);
            return;
        }
        
        for (int i = 0; i < nums.length; i++) {
            if (used[i] == false) {
                if (i != 0 && nums[i] == nums[i-1] && used[i-1] == false) {
                    continue;
                } 
                cur.add(nums[i]);
                used[i] = true;
                helper(res, cur, used, nums);
                used[i] = false;
                cur.remove(cur.size() - 1);
            }
        }
    }
}

 

 

Permutation Sequence

全排列序列

思路：这道题用搜索会超时。放在这是为了与PermutationsI和II对比。这道题可以用O(n)的时间复杂度来求解。

　　　第一位是数字可能是1～n，其中以每个数字打头的有(n-1)!种全排列，因此第一位数应该是 1 ~ n中的第 k / (n-1)!个数，设为a；

　　　接着，问题转化为，1～n中去掉a后，剩下的数中第 k%(n - 1)!个全排列，按照找第一位数的方法再来找第二位数，第三位数..........。

public class Solution {
    public String getPermutation(int n, int k) {
        String res = "";
        List<Integer> nums = new ArrayList<Integer>();
        for (int i = 1; i <= 9; i++) {
            nums.add(i);
        }
        
        //facs[i]表示阶乘i!
        int[] facs = new int[10];
        facs[1] = 1;
        for (int i = 2; i <= 9; i++) {
            facs[i] = facs[i - 1] * i;
        }
        
        int j = 0;   //j表示这一位数是剩余的数的第j个数
        int m = n;   //m表示剩下多少个数
        for (int i = 1; i < n; i++) {
            int fac = facs[m - 1];
            j = k % fac == 0 ? k / fac - 1 : k / fac;
            k = k % fac == 0 ? fac : k % fac;
            res += String.valueOf(nums.get(j));
            nums.remove(j);
            m--;
        }
        res += String.valueOf(nums.get(0));
        return res;
    }
}    

 

 

Subsets II

带重复元素的子集

思路：常规dfs题目,先排序,后搜索.注意如何去重

class Solution {
    /**
     * @param nums: A set of numbers.
     * @return: A list of lists. All valid subsets.
     */
    public ArrayList<ArrayList<Integer>> subsetsWithDup(int[] nums) {
        // write your code here
        Arrays.sort(nums);
        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
        
        helper(res, new ArrayList<Integer>(), nums, 0);
        return res;
        
    }
    
    public void helper(ArrayList<ArrayList<Integer>> res, ArrayList<Integer> cur, 
        int[] nums, int start) {
        
        ArrayList<Integer> tmp = new ArrayList<Integer>(cur);
        res.add(tmp);
        
        for (int i = start; i < nums.length;) {
            cur.add(nums[i]);
            helper(res, cur, nums, i + 1);
            cur.remove(cur.size() - 1);
            
            if (i == nums.length - 1) {
                i++;
            }
            while (i < nums.length - 1 && nums[i++] == nums[i]) {}
            if (i == nums.length - 1 && nums[i-1] == nums[i]) {
                i++;
            }
        }
    }
}

 

 

Six Degrees

六度问题

思路：使用BFS，一层一层搜索，到达目标节点的层数就是答案。

/**
 * Definition for Undirected graph.
 * class UndirectedGraphNode {
 *     int label;
 *     List<UndirectedGraphNode> neighbors;
 *     UndirectedGraphNode(int x) { 
 *         label = x;
 *         neighbors = new ArrayList<UndirectedGraphNode>(); 
 *     }
 * };
 */
public class Solution {
    /**
     * @param graph a list of Undirected graph node
     * @param s, t two Undirected graph nodes
     * @return an integer
     */
    public int sixDegrees(List<UndirectedGraphNode> graph,
                          UndirectedGraphNode s,
                          UndirectedGraphNode t) {
        // Write your code here
        if (s.equals(t)) {
            return 0;
        }
        Set<UndirectedGraphNode> set = new HashSet<UndirectedGraphNode>();
        Queue<UndirectedGraphNode> queue = new LinkedList<UndirectedGraphNode>();
        int layernum = 1;
        
        set.add(s);
        queue.offer(s);
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 1; i <= size; i++) {
                UndirectedGraphNode cur = queue.poll();
                for (int j = 0; j < cur.neighbors.size(); j++) {
                    UndirectedGraphNode next = cur.neighbors.get(j);
                    if (set.contains(next)) {
                        continue;
                    }
                    set.add(next);
                    queue.offer(next);
                    if (next.equals(t)) {
                        return layernum;
                    }
                }
            }
            layernum++;
        }
        return -1;
    }
}

 

 

Topological Sorting

拓扑排序

思路：先找出任意一个没有入边的点。然后显示出该点，并将它及其边从图中删除。然后对图中其余部分应用这种方法。

　　　具体来说：先求每个节点的入度，然后把入度为0的节点放入队列中，出队的时候记录节点并将邻接表中所有节点的入度都减1，如果产生新的入度为0的节点就放入队列。直到队列为空。

/**
 * Definition for Directed graph.
 * class DirectedGraphNode {
 *     int label;
 *     ArrayList<DirectedGraphNode> neighbors;
 *     DirectedGraphNode(int x) { label = x; neighbors = new ArrayList<DirectedGraphNode>(); }
 * };
 */
public class Solution {
    /**
     * @param graph: A list of Directed graph node
     * @return: Any topological order for the given graph.
     */    
    public ArrayList<DirectedGraphNode> topSort(ArrayList<DirectedGraphNode> graph) {
        // write your code here
        Map<DirectedGraphNode, Integer> nodeMap = new HashMap<DirectedGraphNode, Integer>();
        for (DirectedGraphNode node : graph) {
            nodeMap.put(node, 0);
        }
        for (DirectedGraphNode node : graph) {
            for (DirectedGraphNode neighbor : node.neighbors) {
                nodeMap.put(neighbor, nodeMap.get(neighbor) + 1);
            }
        }
        
        Queue<DirectedGraphNode> q = new LinkedList<DirectedGraphNode>();
        for (Map.Entry<DirectedGraphNode, Integer> entry : nodeMap.entrySet()) {
            if (entry.getValue() == 0) {
                q.offer(entry.getKey());
            }
        }
        
        ArrayList<DirectedGraphNode> res = new ArrayList<DirectedGraphNode>();
        while (!q.isEmpty()) {
            DirectedGraphNode cur = q.poll();
            res.add(cur);
            for (DirectedGraphNode neighbor : cur.neighbors) {
                nodeMap.put(neighbor, nodeMap.get(neighbor) - 1);
                if (nodeMap.get(neighbor) == 0) {
                    q.offer(neighbor);
                }
            }
        }
        return res;
        
    }
}

 

 

Word Ladder

单词接龙

思路：第一步：求出每个单词的邻接单词表，即word2neighbors，它的键是dict中每个单词，值是与该单词相差一个单词的所有单词。

　　   第二步：根据邻接单词表，从end开始作bfs遍历，一层一层遍历，直到找到start，返回当前的层数。

public class Solution {
    /**
      * @param start, a string
      * @param end, a string
      * @param dict, a set of string
      * @return an integer
      */
    public int ladderLength(String start, String end, Set<String> dict) {
        // write your code here
        if (start.equals(end)) {
            return 1;
        }
        dict.add(start);
        dict.add(end);
        int len = start.length();
        Map<String, List<String>> word2neighbors = getNeibors(dict, len);
        return bfs(word2neighbors, start, end);
    }
    
    public Map<String, List<String>> getNeibors(Set<String> dict, int len) {
        Map<String, List<String>> word2neighbors = new HashMap<String, List<String>>();
        for (int i = 0; i < len; i++) {
            Map<String, List<String>> rep2words = new HashMap<String, List<String>>();
            for (String word : dict) {
                String rep = word.substring(0, i) + word.substring(i + 1);
                updateMap(rep2words, rep, word);
            }
            for (Map.Entry<String, List<String>> entry : rep2words.entrySet()) {
                List<String> words = entry.getValue();
                for (String word1 : words) {
                    for (String word2 : words) {
                        if (!word1.equals(word2)) {
                            updateMap(word2neighbors, word1, word2);
                        }
                    }
                }
            }
        }
        return word2neighbors;
    }
    
    public int bfs(Map<String, List<String>> word2neighbors, String start, String end) {
        
        Set<String> searched = new HashSet<String>();
        Queue<String> q = new LinkedList<String>();
        q.offer(start);
        searched.add(start);
        int layer = 1;
        
        while(!q.isEmpty()) {
            int size = q.size();
            for (int i = 1; i <= size; i++) {
                String cur = q.poll();
                List<String> neighbors = word2neighbors.get(cur);
                for (String neighbor : neighbors) {
                    if (!searched.contains(neighbor)) {
                        if (neighbor.equals(end)) {
                            return layer + 1;
                        }
                        q.offer(neighbor);
                        searched.add(neighbor);
                    }
                }
            }
            layer++;
        }
        return -1;
    }
    
    public void updateMap(Map<String, List<String>> map, String key, String value) {
        List<String> list = map.get(key);
        if (list == null) {
            list = new ArrayList<String>();
            list.add(value);
            map.put(key, list);
        } else {
            list.add(value);
        }
    }
}

 

 

Word Ladder II

单词接龙II

思路1：本题是搜索题目中的较难题。

　　　第一步：求出每个单词的邻接单词表，即word2neighbors，它的键是dict中每个单词，值是与该单词相差一个单词的所有单词。

　　   第二步：根据邻接单词表，从end开始作bfs遍历，一层一层遍历，直到找到start。并生成prewords表，它的键是遍历过的每个单词，值是在遍历路径上连到该单词的所有的前驱单词；

　　   第三步：根据prewords表，从start开始作dfs遍历，遍历到end时就将路径加入res。

public class Solution {
    /**
      * @param start, a string
      * @param end, a string
      * @param dict, a set of string
      * @return a list of lists of string
      */
    public List<List<String>> findLadders(String start, String end, Set<String> dict) {
        // write your code here  
        dict.add(start);
        dict.add(end);
        int len = start.length();
        Map<String, List<String>> word2neighbors = getNeibors(dict, len);
        Map<String, List<String>> prewords = bfs(word2neighbors, start, end);
        
        List<List<String>> res = new ArrayList<List<String>>();
        List<String> cur = new ArrayList<String>();
        cur.add(start);
        dfs(res, cur, prewords, start, end);
        return res;
        
    }
    
    public Map<String, List<String>> getNeibors(Set<String> dict, int len) {
        Map<String, List<String>> word2neighbors = new HashMap<String, List<String>>();
        for (int i = 0; i < len; i++) {
            Map<String, List<String>> rep2words = new HashMap<String, List<String>>();
            for (String word : dict) {
                String rep = word.substring(0, i) + word.substring(i + 1);
                updateMap(rep2words, rep, word);
            }
            for (Map.Entry<String, List<String>> entry : rep2words.entrySet()) {
                List<String> words = entry.getValue();
                for (String word1 : words) {
                    for (String word2 : words) {
                        if (!word1.equals(word2)) {
                            updateMap(word2neighbors, word1, word2);
                        }
                    }
                }
            }
        }
        return word2neighbors;
    }
    
    public Map<String, List<String>> bfs(Map<String, List<String>> word2neighbors, String start, String end) {
        
        Set<String> searched = new HashSet<String>();
        Map<String, List<String>> prewords = 
            new HashMap<String, List<String>>();
        Queue<String> q = new LinkedList<String>();
        q.offer(end);
        searched.add(end);
        
        boolean find = false;
        while(!q.isEmpty()) {
            int size = q.size();
            Set<String> nextLev = new HashSet<String>();
            for (int i = 1; i <= size; i++) {
                String cur = q.poll();
                List<String> neighbors = word2neighbors.get(cur);
                for (String neighbor : neighbors) {
                    if (!searched.contains(neighbor)) {
                        if (neighbor.equals(start)) {
                            find = true;
                        }
                        if (!nextLev.contains(neighbor)) {
                            q.offer(neighbor);
                            nextLev.add(neighbor);
                        }
                        updateMap(prewords, neighbor, cur);
                    }
                }
            }
            if (find) {
                break;
            }
            searched.addAll(nextLev);
        }
        return prewords;
    }
    
    public void dfs(List<List<String>> res, List<String> cur,
        Map<String, List<String>> prewords, String curstring, String end) {
        
        if (curstring.equals(end)) {
            List<String> tmp = new ArrayList<String>(cur);
            res.add(tmp);
            return;
        }
        
        List<String> words = prewords.get(curstring);
        for (String word : words) {
            cur.add(word);
            dfs(res, cur, prewords, word, end);
            cur.remove(cur.size() - 1);
        }
    } 
    
    public void updateMap(Map<String, List<String>> map, String key, String value) {
        List<String> list = map.get(key);
        if (list == null) {
            list = new ArrayList<String>();
            list.add(value);
            map.put(key, list);
        } else {
            list.add(value);
        }
    }
}

 思路2：（九章算法答案）

　　　第一步：与思路1一样；

　　　第二步：从start做bfs,求出所有节点到start的最短距离distance（相当于位于第几层）。

　　　第三步：从end做dfs，每个节点只往它的邻节点中distance值小1的节点搜索。

public class Solution {
    public List<List<String>> findLadders(String start, String end,
            Set<String> dict) {
        List<List<String>> ladders = new ArrayList<List<String>>();
        Map<String, List<String>> map = new HashMap<String, List<String>>();
        Map<String, Integer> distance = new HashMap<String, Integer>();

        dict.add(start);
        dict.add(end);
 
        bfs(map, distance, start, end, dict);
        
        List<String> path = new ArrayList<String>();
        
        dfs(ladders, path, end, start, distance, map);

        return ladders;
    }

    void dfs(List<List<String>> ladders, List<String> path, String crt,
            String start, Map<String, Integer> distance,
            Map<String, List<String>> map) {
        path.add(crt);
        if (crt.equals(start)) {
            Collections.reverse(path);
            ladders.add(new ArrayList<String>(path));
            Collections.reverse(path);
        } else {
            for (String next : map.get(crt)) {
                if (distance.containsKey(next) && distance.get(crt) == distance.get(next) + 1) { 
                    dfs(ladders, path, next, start, distance, map);
                }
            }           
        }
        path.remove(path.size() - 1);
    }

    void bfs(Map<String, List<String>> map, Map<String, Integer> distance,
            String start, String end, Set<String> dict) {
        Queue<String> q = new LinkedList<String>();
        q.offer(start);
        distance.put(start, 0);
        for (String s : dict) {
            map.put(s, new ArrayList<String>());
        }
        
        while (!q.isEmpty()) {
            String crt = q.poll();

            List<String> nextList = expand(crt, dict);
            for (String next : nextList) {
                map.get(next).add(crt);
                if (!distance.containsKey(next)) {
                    distance.put(next, distance.get(crt) + 1);
                    q.offer(next);
                }
            }
        }
    }

    List<String> expand(String crt, Set<String> dict) {
        List<String> expansion = new ArrayList<String>();

        for (int i = 0; i < crt.length(); i++) {
            for (char ch = 'a'; ch <= 'z'; ch++) {
                if (ch != crt.charAt(i)) {
                    String expanded = crt.substring(0, i) + ch
                            + crt.substring(i + 1);
                    if (dict.contains(expanded)) {
                        expansion.add(expanded);
                    }
                }
            }
        }

        return expansion;
    }
}