#yyds干货盘点# LeetCode 热题 HOT 100：单词搜索

Posted 2022-10-14

题目：

给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。

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不允许被重复使用。

 

示例 1：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"

输出：true

示例 2：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"

输出：true

示例 3：

输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"

输出：false

代码实现：

class Solution 
    public boolean exist(char[][] board, String word) 
        int h = board.length, w = board[0].length;
        boolean[][] visited = new boolean[h][w];
        for (int i = 0; i < h; i++) 
            for (int j = 0; j < w; j++) 
                boolean flag = check(board, visited, i, j, word, 0);
                if (flag) 
                    return true;
                
            
        
        return false;
    

    public boolean check(char[][] board, boolean[][] visited, int i, int j, String s, int k) 
        if (board[i][j] != s.charAt(k)) 
            return false;
         else if (k == s.length() - 1) 
            return true;
        
        visited[i][j] = true;
        int[][] directions = 0, 1, 0, -1, 1, 0, -1, 0;
        boolean result = false;
        for (int[] dir : directions) 
            int newi = i + dir[0], newj = j + dir[1];
            if (newi >= 0 && newi < board.length && newj >= 0 && newj < board[0].length) 
                if (!visited[newi][newj]) 
                    boolean flag = check(board, visited, newi, newj, s, k + 1);
                    if (flag) 
                        result = true;
                        break;
                    
                
            
        
        visited[i][j] = false;
        return result;