为什么这个解决方案在给定的测试用例上给出TLE

Question

问题链接：https://leetcode.com/problems/word-search/

给定一个2D板和一个单词，查找该单词是否存在于网格中。

该字可以由顺序相邻单元的字母构成；相邻的"；单元格是水平或垂直相邻的单元格。同一个字母单元格不能多次使用。

我们不应该使用一个字符两次。

示例：

board =
[
['A','B','C','E'],
['S','F','C','S'],
['A','D','E','E']
]
Given word = "ABCCED", return true.
Given word = "SEE", return true.
Given word = "ABCB", return false.

限制：

board and word consists only of lowercase and uppercase English letters.
1 <= board.length <= 200
1 <= board[i].length <= 200
1 <= word.length <= 10^3

我的逻辑：在函数存在中，每当我在2D数组中找到给定字符串(字符串s(的第一个字符时，我都会在其位置调用DFS，以检查该字符串是否可以形成。

我得到了上述测试用例的TLE

测试用例：

[["a","a","a","a"],["a","a","a","a"],["a","a","a","a"],["a","a","a","a"],["a","a","a","b"]]
"aaaaaaaaaaaaaaaaaaaa"

预期输出：

true

代码：

class Solution {
public:

bool dfs( vector<vector<char>> board , string s, int p ,int i , int j ){
if( p == s.length() ){
return true;
}

if( i < 0 || i >= board.size() || j < 0 || j >= board[i].size() || board[i][j] != s.at(p) ){
return false;
}

char t = board[i][j];
board[i][j] = ' ';

bool res = dfs( board, s, p + 1 , i + 1, j ) | dfs( board, s, p + 1 , i - 1, j ) |
dfs( board, s, p + 1 , i , j + 1 ) | dfs( board, s, p + 1 , i, j - 1 );

board[i][j] = t;

return res;

}

bool exist(vector<vector<char>>& board, string s) {

for( int i = 0; i < board.size(); i++ ){
for( int j = 0; j < board[0].size(); j++ ){
if( board[i][j] == s.at(0) && dfs( board , s , 0, i , j ) ){
return true;
}
}
}
return false;
}
};

提交详细信息：https://leetcode.com/submissions/detail/398643848/

Answer 1

递归函数dfs按值接收板，即在尝试时无法更改板。由于大量复制和无限递归而导致超时。看起来像虫子。