题目
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
题解
class Solution {
public boolean exist(char[][] board, String word) {
int row = board.length;
int col = board[0].length;
boolean res = false;
for(int i = 0; i < row; i++){
for(int j = 0; j < col; j++){
if(board[i][j] == word.charAt(0)){
if(findWay(board, word, i, j, 0, row, col))
return true;
}
}
}
return false;
}
boolean findWay(char[][] board, String word, int i, int j, int index, int row, int col){
if(index == word.length())
return true;
if(i < 0 || i >= row || j < 0 || j >= col || board[i][j] == '#' || board[i][j] != word.charAt(index))
return false;
char tmp = board[i][j]; // 记录下board[i][j]的值以便以该点为首的路径递归结束后还原
board[i][j] = '#';
boolean res = findWay(board, word, i, j - 1, index + 1, row, col)
|| findWay(board, word, i, j + 1, index + 1, row, col)
|| findWay(board, word, i - 1, j, index + 1, row, col)
|| findWay(board, word, i + 1, j, index + 1, row, col);
board[i][j] = tmp; // 删除访问记录，恢复该点之前的状态
return res;
}
}
