LeetCode 79. Word Search

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本文介绍了一种使用深度优先搜索(DFS)算法在二维字符网格中查找特定单词的方法。通过递归方式检查每个网格位置，判断是否能构成目标单词，同一字母单元格在单词构造中不可重复使用。

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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.
Example:board =
[
[‘A’,‘B’,‘C’,‘E’],
[‘S’,‘F’,‘C’,‘S’],
[‘A’,‘D’,‘E’,‘E’]
]

Given word = “ABCCED”, return true.
Given word = “SEE”, return true.
Given word = “ABCB”, return false.

深度优先搜索：使用递归

class Solution {
public:
    bool exist(vector<vector<char>>& board, string word) {
        if(board.size()==0)
            return false;
        h=board.size();
        w=board[0].size();
        for(int i=0;i<h;i++)
            for(int j=0;j<w;j++)
                if(search(board,word,0,i,j))
                    return true;
        return false;
    }
    
    bool search(vector<vector<char>>& board, const string & word,int d,int x,int y){
        if(x<0||x==h||y<0||y==w||word[d]!=board[x][y])
            return false;
        if(d==word.length()-1)
            return true;
        char cur=board[x][y];
        board[x][y]=0;
        bool found=search(board,word,d+1,x+1,y)||
            search(board,word,d+1,x-1,y)||
            search(board,word,d+1,x,y+1)||
            search(board,word,d+1,x,y-1);
        board[x][y]=cur;
        return found;
    }
    
private:
    int w;
    int h;
};