Please design a function to determine whether there is a path that contains all the characters of a string in a matrix. The path can start from any grid in the matrix, and each step can move one grid to the left, right, up, and down in the matrix. If a path passes through a grid of the matrix, then the path cannot enter the grid again. For example, the following 3×4 matrix contains a path for the character string "bfce" (the letters in the path are marked in bold).
[[“a”,“b”,“c”,“e”],
[“s”,“f”,“c”,“s”],
[“a”,“d”,“e”,“e”]]
But the matrix does not contain the path of the string "abfb", because after the first character b of the string occupies the second grid in the first row of the matrix, the path cannot enter this grid again.
Example 1:
Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D", "E","E"]], word = "ABCCED"
Output: true
Example 2:
Input: board = [["a","b"],["c","d"]], word = "abcd"
Output: false
analysis:
The idea of backtracking discusses the recursion of each node after moving up, down, left, and right 通过对上下左右四个边界和当前元素与当前步数对应的值（board[i][j]与word[k]）不同来进行剪枝. code show as below:
class Solution {
public:
bool exist(vector<vector<char>>& board, string word) {
int m = board.size();
int n = board[0].size();
for(int i = 0; i < m; i++){
for(int j = 0; j < n; j++){
if(dfs(board, word, i, j, 0)) return true;
}
}
return false;
}
bool dfs(vector<vector<char>>& board, string word, int i, int j, int k){
if(i > board.size() - 1 || i < 0 || j < 0 || j > board[0].size() -1 || board[i][j] != word[k]){
return false;
}
if(k == word.size() - 1) return true;
board[i][j] = '\0'; // 避免重复访问
bool res = dfs(board, word, i - 1, j, k + 1) || dfs(board, word, i + 1, j, k + 1) ||
dfs(board, word, i, j - 1, k + 1) || dfs(board, word, i, j + 1, k + 1);
board[i][j] = word[k]; // 回溯上一步，还原被访问的位置
return res;
}
};