单词搜索 II——leetcode212(DFS/DFS+字典树)

单词搜索 II

题目：单词搜索 II

给定一个 m x n 二维字符网格 board 和一个单词（字符串）列表 words，找出所有同时在二维网格和字典中出现的单词。

单词必须按照字母顺序，通过 相邻的单元格 内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母在一个单词中不允许被重复使用。

示例：

输入：board = [["o","a","a","n"],["e","t","a","e"],["i","h","k","r"],["i","f","l","v"]], words = ["oath","pea","eat","rain"]
输出：["eat","oath"]

题解

方法一: 单纯的DFS

    会超时

方法二：DFS+前缀剪枝

用set集合存储字典中的前缀

dfs遍历时判断当前字符串在前缀结合中是否存在

如果存在，则继续，否则就退出。

时间复杂度和空间复杂度：10%

class Solution0 {
    private Set<String> results;
    private Set<String> prex;
    private Set<String> wordSet;
    private int[][] nag=new int[][]{{0,1},{0,-1},{-1,0},{1,0}};
    public void dfs(char[][] board, int[][] book, int x, int y, StringBuilder sb){
        if(!prex.contains(sb.toString())) return;       //判断前缀中是否包含sb
        if(wordSet.contains(sb.toString())) {           //判断结果中是否包含sb
            results.add(sb.toString());
        }
        for(int i=0;i<4;i++) {
           int newx=x+nag[i][0], newy=y+nag[i][1];
            if(newx<0 || newx>=board.length || newy<0 || newy>=board[0].length || book[newx][newy]==1) continue;
            book[newx][newy]=1;
            sb.append(board[newx][newy]);
            dfs(board, book, newx, newy, sb);
            sb.deleteCharAt(sb.length()-1);
            book[newx][newy]=0;
        }
    }
    public List<String> findWords(char[][] board, String[] words) {
        int[][] book=new int[board.length][board[0].length];
        results=new HashSet<>();
        prex=new HashSet<>();               //前缀集合
        wordSet=new HashSet<>();            //字典集合

        for(int i=0;i<words.length;i++) {
            for(int j=0;j<words[i].length();j++){
                prex.add(words[i].substring(0, j+1));
            }
            wordSet.add(words[i]);
        }

        StringBuilder sb=new StringBuilder();
        for(int i=0;i<board.length;i++) {
            for(int j=0;j<board[0].length;j++) {
                book[i][j]=1;
                sb.append(board[i][j]);
                dfs(board, book, i, j, sb);
                sb.deleteCharAt(sb.length()-1);
                book[i][j]=0;
            }
        }

        return new ArrayList<>(results);
    }
}

方法三：DFS+字典树

class Solution {
    class Tree
    {
        String word;
        Map<Character, Tree> children;

        public Tree()
        {
            this.word="";
            children=new HashMap<>();
        }
        
        //字典树插入
        public void insert(String word){
            Tree temp=this;
            for(int i=0;i<word.length();i++) {
                char c=word.charAt(i);
                temp.children.putIfAbsent(c, new Tree());
                temp=temp.children.get(c);
            }
            temp.word=word;
        }
    }
    private Set<String> results;
    private int[][] dirs=new int[][]{{1,0},{-1,0},{0,1},{0,-1}};
    public void dfs(char[][] board, Tree root, int x, int y, int[][] book) {
        //字典树：判断board[x][y]是否是当前节点的下一个节点
        if(!root.children.containsKey(board[x][y])) return;
        Tree next=root.children.get(board[x][y]);

        //判断当前节点是否是叶子节点
        if(!next.word.equals("")){
            results.add(next.word);
        }

        //DFS遍历四个方向的节点
        for(int i=0;i<4;i++){
            int newx=x+dirs[i][0], newy=y+dirs[i][1];
            if(newx<0||newx>=board.length|| newy<0|| newy>=board[0].length || book[newx][newy]==1) continue;
            book[newx][newy]=1;
            dfs(board, next, newx, newy, book);
            book[newx][newy]=0;
        }
    }
    public List<String> findWords(char[][] board, String[] words) {
        results=new HashSet<>();
        int book[][]=new int[board.length][board[0].length];
        Tree root=new Tree();
        for (String word : words) {
            root.insert(word);
        }

        StringBuilder sb=new StringBuilder();
        for(int i=0;i<board.length;i++){
            for(int j=0;j<board[0].length;j++) {
                book[i][j]=1;
                dfs(board, root, i, j, book);
                book[i][j]=0;
            }
        }
        return new ArrayList<>(results);
    }
}