做题博客链接
https://blog.csdn.net/qq_43349112/article/details/108542248
题目链接
https://leetcode-cn.com/problems/ju-zhen-zhong-de-lu-jing-lcof/
描述
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。
提示：
1 <= board.length <= 200
1 <= board[i].length <= 200
示例
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
初始代码模板
class Solution {
public boolean exist(char[][] board, String word) {
}
}
代码
class Solution {
public boolean exist(char[][] board, String word) {
if (board.length * board[0].length < word.length()) {
return false;
}
char[] words = word.toCharArray();
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
if (dfs(board, i, j, words, 0)) {
return true;
}
}
}
return false;
}
private boolean dfs(char[][] board, int i, int j, char[] words, int idx) {
if (idx == words.length) {
return true;
}
if (i < 0 || i >= board.length || j < 0 || j >= board[0].length || board[i][j] != words[idx]) {
return false;
}
board[i][j] = '\0';
int nextIdx = idx + 1;
boolean flag = dfs(board, i, j + 1, words, nextIdx) ||
dfs(board, i, j - 1, words, nextIdx) ||
dfs(board, i + 1, j, words, nextIdx) ||
dfs(board, i - 1, j, words, nextIdx);
board[i][j] = words[idx];
return flag;
}
}