# C++实现LeetCode(79.词语搜索)

[LeetCode] 79. Word Search 词语搜索 Given a 2D board and a word, find if the word exists in the grid.

## [LeetCode] 79. Word Search 词语搜索

Given a 2D board and a word, find if the word exists in the grid.

The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

For example,

Given board =

mXUQMUHPjr

[

["ABCE"],

["SFCS"],

]

word = "ABCCED", -> returns true,

word = "SEE", -> returns true,

word = "ABCB", -> returns false.

```class Solution {
public:
bool exist(vector<vectmXUQMUHPjror<char>>& board, string word) {
if (board.empty() || board[0].empty()) return false;
int m = board.size(), n = board[0].size();
vector<vector<bool>> visited(m, vec编程客栈tor<bool>(n));
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (search(board, word, http://www.cppcns.com0, i, j, visited)) return true;
}
}
return false;
}
bool search(vector<vector<char>>& board, string word, int idx, int i, int j, vector<vector<bool>>& visited) {
if (idx == word.size()) return true;
int m = board.size(), n = board[0].size();
if (i < 0 || j < 0 || i >= m || j >= n || visited[i][j] || board[i][j] != word[idx]) return false;
visited[i][j] = true;
bool res = search(board, word, idx + 1, i - 1, j, visited)
|| search(board, word, idx + 1, i + 1, j, visited)
|| search(board, word, idx + 1, i, j - 1, visited)
|| search(board, word, idx + 1, i, j + 1, visited);
visited[i][j] = false;
return res;
}
};```

```class Solution {
public:
bool exist(vector<vector<char>>& board, string word) {
if (board.empty() || board[0].empty()) return false;
int m = board.size(), n = board[www.cppcns.com0].size();
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (search(board, word, 0, i, j)) return true;
}
}
return false;
}
bool search(vector<vector<char>>& board, string word, int idx, int i, int j) {
if (idx == word.size()) return true;
int m = board.size(), n = board[0].size();
if (i < 0 || j < 0 || i >= m || j >= n || board[i][j] != word[idx]) return false;
char c = board[i][j];
board[i][j] = '#';
bool res = search(board, word, idx + 1, i - 1, j)
|| search(board, word, idx + 1, i + 1, j)
|| search(board, word, idx + 1, i, j - 1)
|| search(board, word, idx + 1, i, j + 1);
board[i][j] = c;
return res;
}
};```