[LeetCode] 79. Word Search

[grandyang]

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Given an m x n grid of characters board and a string word, return true if word exists in the grid.

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

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","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
Output: true

Example 3:

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
Output: false

Constraints:

• m == board.length
• n = board[i].length
• 1 <= m, n <= 6
• 1 <= word.length <= 15
• board and word consists of only lowercase and uppercase English letters.

Follow up: Could you use search pruning to make your solution faster with a larger board?

这道题是典型的深度优先遍历 DFS 的应用，原二维数组就像是一个迷宫，可以上下左右四个方向行走，这里以二维数组中每一个数都作为起点和给定字符串做匹配，还需要一个和原数组等大小的 visited 数组，是 bool 型的，用来记录当前位置是否已经被访问过，因为题目要求一个 cell 只能被访问一次。如果二维数组 board 的当前字符和目标字符串 word 对应的字符相等，则对其上下左右四个邻字符分别调用 DFS 的递归函数，只要有一个返回 true，那么就表示可以找到对应的字符串，否则就不能找到，具体看代码实现如下：

解法一：

class Solution {
public:
    bool exist(vector<vector<char>>& board, string word) {
        int m = board.size(), n = board[0].size();
        vector<vector<bool>> visited(m, vector<bool>(n));
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (search(board, word, 0, 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;
    }
};

我们还可以不用 visited 数组，直接对 board 数组进行修改，将其遍历过的位置改为井号，记得递归调用完后需要恢复之前的状态，参见代码如下：

解法二：

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 (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;
    }
};

Github 同步地址：

#79

类似题目：

Word Search II

参考资料：

https://leetcode.com/problems/word-search/

https://leetcode.com/problems/word-search/discuss/27658/Accepted-very-short-Java-solution.-No-additional-space.

https://leetcode.com/problems/word-search/discuss/27829/C++-backtracking-solution-without-extra-data-structure

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[lei-hsia]

博主好，对最后的board[i]][j] = c这个回溯，你觉得什么时候应该写啊？
我看比如像200. Number of Islands纯粹的dfs就没有，但是像这种backtracking就要回溯。是不是因为那道题只是“相邻的island”全部0变成1，只要变成1就是属于当前island后面就不用再检验；而这里word的检验，如果不相等后面还要回到原点进行检验？
不知道我说清楚没有，我感觉这种区别很微妙

[grandyang]

差不多就是这个意思，200题里岛屿1的顺序并没有关系，而这里的单词是按某个顺序连起来的～