# [79\. Word Search](https://leetcode.com/problems/word-search/)

- 在矩陣裡找字存不存在。
- Backtracking 終止條件有兩個：return false 的狀況跟順利搜尋完的狀況。

:::spoiler Solution

class Solution {
public:
    bool exist(vector<vector<char>>& board, string word)
    {
        int m = board.size(), n = board[0].size();

        for (int i = 0; i < m; ++i)
        {
            for (int j = 0; j < n; ++j)
            {
                if (board[i][j] == word[0] && backtracking(board, word, i, j, 0))
                {
                    return true;
                }
            }
        }
        return false;
        
    }

    bool backtracking(vector<vector<char>>& board, string word, int i, int j, int start)
    {
        int m = board.size(), n = board[0].size();

        if (i < 0 || j < 0 || i >= m || j >= n || board[i][j] != word[start])
        {
            return false;
        }

        if (start == word.size() - 1) return true;

        char temp = board[i][j];

        board[i][j] = '#';

        bool res = backtracking(board, word, i + 1, j, start + 1) || \
                   backtracking(board, word, i - 1, j, start + 1) || \
                   backtracking(board, word, i, j + 1, start + 1) || \
                   backtracking(board, word, i, j - 1, start + 1);

        board[i][j] = temp;
        return res;
    }
};

- 時間複雜度：$O(n \cdot 3^L)$
- 空間複雜度：$O(L)$
:::