[LeetCode] Word Search

最新推荐文章于 2024-10-24 23:48:44 发布

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本文介绍了一个算法，用于在二维字符网格中查找指定的单词。该算法通过深度优先搜索（DFS）遍历网格，检查单词是否能通过相邻字母构成。文章详细解释了实现过程及注意事项。

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 =

[
  ["ABCE"],
  ["SFCS"],
  ["ADEE"]
]

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,

word = "ABCB", -> returns false.

class Solution {
public:
    bool ans = false;
    bool visited[100][100];
    int step[4][2] = {{-1,0},{1,0},{0,-1},{0,1}};
    string s;
    bool exist(vector<vector<char> > &board, string word) {
        int m = board.size(),n = board[0].size();
        memset(visited,0,sizeof(visited));
        s = word;
        
        for(int i = 0;i < m && !ans;i ++){
            for(int j = 0;j < n && !ans;j ++){
                if(board[i][j] == s[0]){
                    visited[i][j] = true;
                    dfs(1,board,i,j);
                    visited[i][j] = false;
                }
            }
        }
        return ans;
    }
    void dfs(int dep,vector<vector<char> > &board,int x,int y){
        if(dep == s.size()){
            ans = true;
            return;
        }
        for(int i = 0;i < 4 && !ans;i ++){
            int tx = x + step[i][0],ty = y + step[i][1];
            if(tx >= 0 && tx < board.size() && ty >= 0 && ty < board[0].size() && !visited[tx][ty]){
                if(board[tx][ty] == s[dep]){
                    visited[tx][ty] = true;
                    dfs(dep + 1,board,tx,ty);
                    visited[tx][ty] = false;
                }
            }
        }
    }
};