LeetCode 79.单词搜索

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

 

提示：

• m == board.length
• n = board[i].length
• 1 <= m, n <= 6
• 1 <= word.length <= 15
• board 和 word 仅由大小写英文字母组成

 

进阶：你可以使用搜索剪枝的技术来优化解决方案，使其在 board 更大的情况下可以更快解决问题？

Related Topics

数组

回溯

矩阵

👍 988

👎 0

class Solution {
    public boolean exist(char[][] board, String word) {
        int n = board.length;
        int m = board[0].length;
        char words[] = word.toCharArray();
        for(int i = 0; i < n; i++){
            for(int j = 0; j < m; j++){
                if(dfs(board,words,i,j,0)) return true;
            }
        }
        return false;
    }
    public boolean dfs(char[][] board, char[] word, int i, int j, int index){
        if(i<0 || i>=board.length || j<0 || j>=board[0].length || board[i][j]!=word[index]){
            return false;
        }
        if(index == word.length-1) return true;
        
        board[i][j] = '\0'; //把字母删掉
        boolean res = dfs(board,word,i+1,j,index+1) 
            || dfs(board,word,i-1,j,index+1)
            || dfs(board,word,i,j+1,index+1) 
            || dfs(board,word,i,j-1,index+1);
        board[i][j] = word[index];
        return res;
    }
}