leetcode动态规划笔记一---一维DP

动态规划

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告别动态规划，连刷 40 道题，我总结了这些套路，看不懂你打我 - 知乎
 北美算法面试的题目分类，按类型和规律刷题
 九章的DP分类
 九章DP分类
 重点250

题目分类

一维dp

House Robber ：求最大最小值
House Robber II：求最大最小值
House Robber III：求最大最小值
Decode Ways : 求方法总数
Word Break : 求是否可行
Maximum Product Subarray : 求最大最小值
Ugly Number II : 这道题不太像是dp
Is Subsquence :求是否可行
413. Arithmetic Slices ：求总数

矩阵型DP

House Robber : 一维dp

方法一：自定向下 + memo

这种方法是从问题整体出发，向子问题方向分解求值，同时将子问题结果缓存下来。
一般来说这种方法在分析出，子问题与父问题关系后，能直接想到。
注：如果去掉memo，那就是递归解问题的方法了，可以用在子问题解不重叠的场景。

class Solution {
    Map<Integer,Integer> m = new HashMap<>();
    int helper(int[] nums, int right){
        if(m.containsKey(right)){
            return m.get(right);
        }
        
        if(right == 0){
            m.put(right, nums[0]);
            return nums[0];
        }
        else if(right == 1){
            int i = Math.max(nums[0], nums[1]);
            m.put(right, i);
            return i;
        }
        
        int i = Math.max(helper(nums, right - 2) + nums[right], helper(nums, right-1));
        m.put(right, i);
        return i;
    }
    public int rob(int[] nums) {
        if(nums.length == 0) return 0;
        return helper(nums, nums.length - 1);
    }
}

方法二：自底向上 + memo

由小问题向大问题推导，这个思路就必须有memo缓存了，不过可以考虑进一步优化缓存.
这个效率已经要比方法一快了.

class Solution {
    public int rob(int[] nums) {
        if(nums.length == 0) return 0;
        
        int[] memo = new int[nums.length];
        for(int i=0; i<nums.length; i++){
            if(i == 0 ) memo[i] = nums[i];
            else if(i == 1){
                memo[i] = Math.max(nums[0], nums[1]);
            }
            else{
                memo[i] = Math.max(memo[i-1], memo[i-2] + nums[i]);
            }
        }
        
        return memo[nums.length - 1];
    }
}

方法三：自底向上 + memo optimised

class Solution {
    public int rob(int[] nums) {
        if(nums.length == 0) return 0;
        if(nums.length == 1) return nums[0];
        
        int n_1 = nums[0], n_2 = 0;
        for(int i=1; i<nums.length; i++){
            int t = Math.max(n_1, n_2 + nums[i]);
            n_2 = n_1; n_1 = t;
        }
        
        return n_1;
    }
}

House Robber II : 一维dp

方法：同上

class Solution {
    public int rob(int[] nums) {
        if(nums.length == 0) return 0;
        if(nums.length == 1) return nums[0];
        if(nums.length == 2) return Math.max(nums[0], nums[1]);
        int x = 0;
        // 1. drop the last num
         int n2 = nums[0], n1 = Math.max(nums[0], nums[1]);
        for(int i = 2; i < nums.length -1; i ++){
            int t = Math.max(n2 + nums[i], n1);
            n2 = n1;
            n1 = t;
        }
        x = n1;
        
        n2 = 0;
        n1 = nums[1];
        for(int i = 2; i < nums.length; i ++){
            int t = Math.max(n2 + nums[i], n1);
            n2 = n1;
            n1 = t;
        }
        
        return Math.max(x, n1);
    }
}

House Robber III:一维DP

方法一：自顶向下 + memo
这种方式写起来容易出错，效率也比较低

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    Map<TreeNode, Integer> m = new HashMap<TreeNode, Integer>();
    Map<TreeNode, Integer> m2 = new HashMap<TreeNode, Integer>();
    
    int helper(TreeNode t, boolean parentUsed){
        if(t == null) return 0;      
        
        if(parentUsed){
            if(m.containsKey(t)){
                return m.get(t);
            }            
            int x = helper(t.left, false) + helper(t.right, false);
            m.put(t, x);
            return x;
        }
        else{
            if(m2.containsKey(t)){
                return m2.get(t);
            }  
            
            int notUse = helper(t.left, false) + helper(t.right, false);
            
            int use = t.val + helper(t.left, true) + helper(t.right, true);
            
            int x = Math.max(use, notUse);
            
            m2.put(t, x);
            return x;
        }
    }
    public int rob(TreeNode root) {
        if(root == null) return 0;
        
        return Math.max(root.val + 
                        helper(root.left, true) + helper(root.right,true), 
                        helper(root.left, false) + helper(root.right, false));
    }
}

方法二：自顶向下 + memo
虽然也是自顶向下，下面的方法就简洁很多；特别注意标important那行容易出错

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    
    int[] helper(TreeNode t){
        int[] re = new int[2];
        if(null == t){
            re[0] = 0;
            re[1] = 0;
            return re;
        }
        
        int[] left = helper(t.left);
        int[] right = helper(t.right);
        
        re[0] = t.val + left[1] + right[1];
        re[1] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]); // important
        return re;
    }
    public int rob(TreeNode root) {
        if(root == null) return 0;
        
        int[] re = helper(root);
        
        return Math.max(re[0], re[1]);
    }
}

Decode Ways : 一维DP，求方法总数

解法一：自顶向下，递归 + no memo, time limit exceed
这道题真有点不太好想

class Solution {
    int helper(char[] str, int right){
        int sum1 = 0, sum2 = 0;

        if(right < 0) return 1;
        if(right == 0){
            return (str[right] != '0') ? 1 : 0;
        }
        
        if(str[right] != '0'){
            sum1 = helper(str, right-1);
        }
        
        if(str[right-1] != '0'){
            Integer x = Integer.valueOf(new String(str, right - 1, 2));
            if(x >= 1 && x <= 26){
                sum2 = helper(str, right-2);
            }
        }
        
        return sum1 + sum2;
    }
    
    public int numDecodings(String s) {
        return helper(s.toCharArray(), s.length() - 1);
    }
}

解法二：自顶向下，递归 + memo

class Solution {
    Map<Integer, Integer> m = new HashMap<>();
    
    int helper(char[] str, int right){
        int sum1 = 0, sum2 = 0;
        
        if(m.containsKey(right)){
            return m.get(right);
        }
        
        if(right < 0) return 1;
        if(right == 0){
            return (str[right] != '0') ? 1 : 0;
        }
        
        if(str[right] != '0'){
            sum1 = helper(str, right-1);
        }
        
        if(str[right-1] != '0'){
            Integer x = Integer.valueOf(new String(str, right - 1, 2));
            if(x >= 1 && x <= 26){
                sum2 = helper(str, right-2);
            }
        }
        
        m.put(right, sum1+sum2);
        return sum1 + sum2;
    }
    
    public int numDecodings(String s) {
        return helper(s.toCharArray(), s.length() - 1);
    }
}

方法三：自底向上 + 加 O(n) memo

class Solution {    
    public int numDecodings(String s) {
        if(s.length() == 0 || s.charAt(0) == '0') return 0;
        
        int[] memo = new int[s.length() + 1];
        memo[0] = 1; memo[1] = 1;
        
        char[] str = s.toCharArray();
        
        for(int i=1; i<s.length(); i++){
            
            int t1 = 0, t2 = 0;
            if(str[i] != '0'){
                t1 = memo[i];
            }
            
            if(str[i-1] != '0'){
                Integer x = Integer.valueOf(new String(str, i-1, 2));
                if(x >=  10 && x <= 26){
                    t2 = memo[i-1];
                }
            }
            
            memo[i+1] = t1 + t2;
        }
        
        return memo[s.length()];
    }
}

Word Break : 一维DP，求是否可行

方法一 ：自顶向下
递归 + memo

class Solution {
    Map<Integer, Boolean> m = new HashMap<>();
    
    boolean helper(char[] str, int right, Set<String> dict){
        if(right >= str.length){
            return true;
        }
        
        if(m.containsKey(right)){
            return m.get(right);
        }
        
        for(int j = 1; j < str.length - right + 1; j++){
            if(dict.contains(new String(str, right, j))){
                if(helper(str, right + j, dict)){
                    return true;
                }
            }
        }
        
        m.put(right, false);
        return false;
    }
    
    public boolean wordBreak(String s, List<String> wordDict) {
        Set<String> dict = new HashSet<>();
        
        for(String r : wordDict){
            dict.add(r);
        }
        
        return helper(s.toCharArray(), 0, dict);
    }
}

方法二：自底向上
递推 + memo
（ps：这道题并不像经典dp，它当前状态不仅依赖于上阶段的状态，而是依赖于过去所有阶段的状态，看看grandyang的解说）

class Solution {
      public boolean wordBreak(String s, List<String> wordDict) {
        Set<String> dict = new HashSet<>();
        int maxi = 0;
        for(String r : wordDict){
            dict.add(r);
            maxi = Math.max(maxi, r.length()); 
        }
        
        char[] arr = s.toCharArray();
        boolean[] can = new boolean[s.length() + 1];
        can[0] = true;  
        
        for(int i = 0; i < s.length(); i++){
            
            boolean c = false;
            for(int j = i; j >= 0 && j >= (i - maxi); j--){ // j >= (i - maxi)是优化部分
                if(can[j] && dict.contains(new String(arr, j, i+1 - j))){
                    c = true;break;
                }
            }
            can[i+1] = c;
        }
        
        return can[s.length()];
    }
}

Maximum Product Subarray : 求最大最小值

方法一 ：自底向上
递推 + memo
这道题稍微加了一个弯，因为有正负数，我们必须记录上一阶段求得的最大最小值，否则，就不能直接根据上一阶段的结果，直接推导本阶段值。

class Solution {
    class Maxi{
        int maxi;
        int mini;
        
        Maxi(){
            maxi = 1;
            mini = 1;
        }
    }
    
    public int maxProduct(int[] nums) {

        Maxi[] memo = new Maxi[nums.length + 1];
        memo[0] = new Maxi();
        
        for(int i=0; i<nums.length; i++){
            memo[i+1] = new Maxi();
            
            if(nums[i] > 0){
                int x = Math.max(memo[i].maxi, memo[i].mini);
                memo[i+1].maxi = (x > 0) ? x * nums[i] : nums[i];
                
                int y = Math.min(memo[i].maxi, memo[i].mini);
                memo[i+1].mini = (y <= 0) ? y * nums[i] : nums[i];
            }
            else if(nums[i] < 0){
                int x = Math.min(memo[i].maxi, memo[i].mini);
                memo[i+1].maxi = (x <= 0) ? x * nums[i] : nums[i];
                
                int y = Math.max(memo[i].maxi, memo[i].mini);
                memo[i+1].mini = (y > 0) ? y * nums[i] : nums[i];
            }
            else {
                memo[i+1].maxi = memo[i+1].mini = 0;
            }
        }
        
        int maxi = Integer.MIN_VALUE;
        for(int i = 1; i < nums.length + 1; i++){
            maxi = Math.max(maxi, memo[i].maxi);
        }
        return maxi;
    }
}

Ugly Number II

方法一：递推
这个解法的归并排序很经典

class Solution {       
    public int nthUglyNumber(int n) {
        if(n <= 5) return n;
        
        List<Integer> li = new ArrayList<>();
        li.add(1);
        
        int idx2 = 0;
        int idx3 = 0;
        int idx5 = 0;
        while(li.size() < n){
            int v2 = 2 * li.get(idx2);
            int v3 = 3 * li.get(idx3);
            int v5 = 5 * li.get(idx5);
            
            int mini = Math.min(v2, Math.min(v3, v5));
            li.add(mini);
            
            if(mini == v2) idx2++;
            if(mini == v3) idx3++;
            if(mini == v5) idx5++;
            
        }
        
        return li.get(n-1);
    }
}

方法二 ：最小堆
这种写法效率低点，但是是个思路；另外，需搞懂1. Comparator这种写法是什么语法； 2. Priority, Iterator的用法

class Solution {       
    public int nthUglyNumber(int n) {
        if(n <= 5) return n;
        
        PriorityQueue<Long> q = new PriorityQueue<>(
            new Comparator<Long>(){
                public int compare(Long a, Long b){
                    if(a < b){
                        return -1;
                    }
                    else if(a > b){
                        return 1;
                    }
                    else {
                        return 0;
                    }
                }
            });
        
        q.add((long)1);
        
        for(int i = 1; i < n; i++){
            long x = q.remove();
            
            Iterator<Long> it = q.iterator();
            while(it.hasNext()){
                if(x == it.next()) it.remove();
            }
            
            q.add( (long)(x * 2) );
            q.add( (long)(x * 3) );
            q.add( (long)(x * 5) );
        }
        
        return q.peek().intValue();
    }
}

Is Subsquence

方法一：自底向上，递推 + memo
可以将O(N)空间优化为常数

class Solution {
    public boolean isSubsequence(String s, String t) {
        int lenS = s.length();
        int lenT = t.length();
        if(lenS > lenT) return false;
        
        int memo = new int[lenS + 1];
        memo[0] = -1;
        
        char[] arrS = s.toCharArray();
        char[] arrT = t.toCharArray();
        
        for(int i = 0; i < lenS; i ++){
            memo[i + 1] = lenT;
            for(int j = memo[i] + 1; j < lenT; j ++){
                if(arrT[j] == arrS[i]){
                    memo[i + 1] = j;
                    break;
                }
            }
        }
        
        return memo[lenS] != lenT;
    }
}

413. Arithmetic Slices

方法一 ：参考grandyang
方法二 : memo + 递推
dp[i]表示以位置i上的数字结尾的Arithmetic Slice的数目
这道题主要要搞清楚，为啥dp[i] = dp[i-1] + 1

class Solution {
    public int numberOfArithmeticSlices(int[] A) {
        
        if( A.length < 3 ) {
            return 0;
        }
        
        int cnt = 0;
        int[] dp = new int[A.length];
        
        for( int i=2; i < A.length; i++ ) {
            if(A[i] - A[i-1] == A[i-1] - A[i-2]) {
                dp[i] = dp[i-1] + 1;
            }
            
            cnt += dp[i];
        }
        
        return cnt;
    }
}

posted @ 2019-11-28 15:22 holidays 阅读(551) 评论(0) 编辑收藏举报

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