【剑指Offer】面试题12. 矩阵中的路径(DFS)

请设计一个函数，用来判断在一个矩阵中是否存在一条包含某字符串所有字符的路径。路径可以从矩阵中的任意一格开始，每一步可以在矩阵中向左、右、上、下移动一格。如果一条路径经过了矩阵的某一格，那么该路径不能再次进入该格子。例如，在下面的3×4的矩阵中包含一条字符串“bfce”的路径（路径中的字母用加粗标出）。

[["a","b","c","e"],

["s","f","c","s"],

["a","d","e","e"]]

但矩阵中不包含字符串“abfb”的路径，因为字符串的第一个字符b占据了矩阵中的第一行第二个格子之后，路径不能再次进入这个格子。

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","d"]], word = "abcd"

输出：false

提示：

1 <= board.length <= 200

1 <= board[i].length <= 200

class Solution {
    public boolean exist(char[][] board, String word) {
        char[] words = word.toCharArray();
        for(int i = 0; i < board.length; i++){
            for(int j = 0; j < board[0].length; j++){
                if(dfs(board, words, i, j, 0)) return true;
            }
        }
        return false;
    }
    public boolean dfs(char[][] board, char[] words, int i, int j, int k){
        if(i >= board.length || i < 0 || j < 0 || j >= board[0].length || board[i][j] != words[k])
            return false;
        if(k == words.length - 1) return true;
        char temp = board[i][j];
        board[i][j] = ‘#‘;
        boolean res = dfs(board, words, i + 1, j, k + 1) || dfs(board, words, i - 1, j, k + 1)
                    || dfs(board, words, i, j + 1, k + 1) || dfs(board, words, i, j - 1, k + 1);
        board[i][j] = temp;
        return res;
    }
}

原文地址：https://www.cnblogs.com/whisperbb/p/12623944.html