【LeetCode】Word Search 解题报告

【题目】

Given a 2D board and a word, find if the word exists in the grid.

The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

For example,

Given board =

[
  ["ABCE"],
  ["SFCS"],
  ["ADEE"]
]

word =

"ABCCED"

,
-> returns

true

,

word =

"SEE"

,
-> returns

true

,

word =

"ABCB"

,
-> returns

false

.

【解析】

二维数组的字符矩阵，给定一个字符串word，问能不能在字符矩阵中找到这个word，字符矩阵中的字符可以上下左右四个方向形成字符串。

直接用DFS回溯就好。

【Java代码】

public class Solution {
    public boolean exist(char[][] board, String word) {
        int m = board.length;
        int n = board[0].length;
        boolean[][] visited = new boolean[m]
;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (dfs(board, visited, i, j, word, 0)) return true;
            }
        }
        return false;
    }
    
    public boolean dfs(char[][] board, boolean[][] visited, int x, int y, String word, int k) {
        if (k == word.length()) return true;
        if (x < 0 || x >= board.length || y < 0 || y >= board[0].length) return false;
        if (visited[x][y]) return false;
        if (board[x][y] != word.charAt(k)) return false;
        
        visited[x][y] = true;
        if (dfs(board, visited, x - 1, y, word, k + 1)) return true;
        if (dfs(board, visited, x + 1, y, word, k + 1)) return true;
        if (dfs(board, visited, x, y - 1, word, k + 1)) return true;
        if (dfs(board, visited, x, y + 1, word, k + 1)) return true;
        visited[x][y] = false;
        
        return false;
    }
}