Leetcode 79. Word Search

最新推荐文章于 2025-09-05 23:16:08 发布

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最新推荐文章于 2025-09-05 23:16:08 发布

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本文介绍了一个使用递归算法解决二维网格中查找特定单词的问题。通过修改字母标记避免重复使用，并利用递归方式检查所有可能路径，确保单词按顺序出现在相邻单元格中。

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 =

[
[‘A’,’B’,’C’,’E’],
[‘S’,’F’,’C’,’S’],
[‘A’,’D’,’E’,’E’]
]

word = “ABCCED”, -> returns true,
word = “SEE”, -> returns true,
word = “ABCB”, -> returns false.
递归

class Solution {
public:
    bool exist(vector<vector<char>>& board, string word) {
        size_t nr = board.size();
        if (nr == 0) return false;
        size_t nc = board[0].size();       
        size_t r = 0;
        while (r != nr) {
            for (size_t c = 0; c != board[r].size(); ++c) {
                if (board[r][c] == word[0]) {
                    board[r][c] = '\0';
                    bool ret = findChar(board, r, c, string(word.begin() + 1, word.end()));
                    if (ret) return ret;
                    board[r][c] = word[0];
                }
            }
            ++r;
        }   
        return false;
    }

private:
    bool findChar(vector<vector<char> >& board, size_t r, size_t c, string word) { // 以（r, c)为中心寻找word[0];
        if (word.size() == 0) return true;
        vector<pair<size_t, size_t> > rang;
        if (c != board[0].size() - 1) rang.push_back(make_pair(r, c + 1));
        if (c != 0) rang.push_back(make_pair(r, c - 1));
        if (r != board.size() - 1) rang.push_back(make_pair(r + 1, c));
        if (r != 0) rang.push_back(make_pair(r - 1, c));
        for (auto p : rang) {
            if (board[p.first][p.second] == word[0]) {
                board[p.first][p.second] = '\0';
                bool ret = findChar(board, p.first, p.second, string(word.begin() + 1, word.end()));
                if (ret) return ret;
                board[p.first][p.second] = word[0];
            }
        }
        return false;
    }
};

参考后

class Solution {
public:
    bool exist(vector<vector<char>>& board, string word) {
        size_t nr = board.size();
        if (nr == 0) return false;
        size_t nc = board[0].size();       
        size_t r = 0;
        while (r != nr) {
            for (size_t c = 0; c != board[r].size(); ++c) {
                bool ret = findChar(board, r, c, word);
                if (ret) return ret;
            }
            ++r;
        }   
        return false;
    }

private:
    bool findChar(vector<vector<char> >& board, int r, int c, string word) { // 以（r, c)为中心寻找word[0];
        if (word.size() == 0) return true;
        if (r < 0 || r >= board.size() || c < 0 || c >= board[0].size())   return false;
        if (board[r][c] == word[0]) {
            board[r][c] = '\0';
            bool ret = (findChar(board, r, c - 1, string(word.begin() + 1, word.end()))
                    || findChar(board, r, c + 1, string(word.begin() + 1, word.end()))
                    || findChar(board, r - 1, c, string(word.begin() + 1, word.end()))   
                    || findChar(board, r + 1, c, string(word.begin() + 1, word.end())));
            if (ret) return ret;
            board[r][c] = word[0];
        }
        return false;
    }
};