Word Search（leetcode）

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本文介绍了一道LeetCode上的经典题目——单词搜索的解决方案。该题要求在二维字符网格中查找给定单词是否存在，单词由相邻单元格的字母构成且每个单元格只能使用一次。文章提供了详细的递归算法实现，通过深度优先搜索来遍历所有可能路径。

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Word Search（leetcode）

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.

代码：

class Solution {
public:
    bool exist(vector<vector<char>>& board, string word) {
        int x = board.size();
        int y = board[0].size();
        bool result = false;
        for(int i = 0; i < x; i++) {
            for (int j = 0; j < y; j++) {
                result = isFound(board, i, j, word.c_str());
                if (result)
                    return true;
            }
        }
        return result;
    }
    bool isFound(vector<vector<char>>& board, int i, int j, const char* word) {
        int x = board.size();
        int y = board[0].size();
        if (*(word) == '\0')
            return true;
        if (i < 0 || j < 0 || i >= x || j >= y || board[i][j] != *word)
            return false;
        char c = board[i][j];
        board[i][j] = '#';
        if (isFound(board, i-1, j, word+1) ||
            isFound(board, i, j+1, word+1) ||
            isFound(board, i+1, j, word+1) ||
            isFound(board, i, j-1, word+1))
            return true;
        board[i][j] = c;
        return false;
    }
};