LeetCode OJ Word Search

最新推荐文章于 2025-09-08 16:25:29 发布

Nightonke

最新推荐文章于 2025-09-08 16:25:29 发布

阅读量449

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode 文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/u012925008/article/details/44518695

leetcode 专栏收录该内容

219 篇文章

订阅专栏

本文介绍了一种用于在二维网格中查找指定单词的算法。通过相邻单元格的顺序连接构建单词，确保同一字母单元格仅使用一次。具体实现了搜索函数，包括初始化变量、方向数组、标记矩阵和访问矩阵等，确保高效地定位目标单词。

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 exist(vector<vector<char> > &board, string word) {
        if (board.size() == 0 || (board.size() * board[0].size() < word.size())) {
            return false;
        }
        dir[0][0] = -1;
        dir[0][1] = 0;
        dir[1][0] = 1;
        dir[1][1] = 0;
        dir[2][0] = 0;
        dir[2][1] = 1;
        dir[3][0] = 0;
        dir[3][1] = -1;
        ans = false;
        
        t_length = word.size();
        target = new char[word.size()];
        for (int i = 0; i < word.size(); i++) target[i] = word[i];
        
        h = board.size();
        w = board[0].size();
        
        mapp = new char*[h];
        for (int i = 0; i < h; i++) mapp[i] = new char[w];
        
        vis = new short*[h];
        for (int i = 0; i < h; i++) vis[i] = new short[w];
        
        
        for (int ii = 0; ii < h; ii++) {
            for (int jj = 0; jj < w; jj++) {
                mapp[ii][jj] = board[ii][jj];
                vis[ii][jj] = 0;
            }
        }
        
        for (int i = 0; i < h; i++) {
            for (int j = 0; j < w; j++) {
                if (board[i][j] == word[0]) {
                    
                    vis[i][j] = 1;
                    search(1, i, j);
                    if (ans) {
                        return ans;
                    }
                    vis[i][j] = 0;
                }
            }
        }
        
        for (int i = 0; i < h; i++) delete []mapp[i];
        delete []mapp;
        
        for (int i = 0; i < h; i++) delete []vis[i];
        delete []vis;
        
        return ans;
    }
    void search(int pos, int i, int j) {
        if (pos == t_length) {
            ans = true;
            return;
        }
        for (int k = 0; k < 4 && ans == false; k++) {
            if (0 <= i + dir[k][0] && i + dir[k][0] < h && 0 <= j + dir[k][1] && j + dir[k][1] < w && (!vis[i + dir[k][0]][j + dir[k][1]]) && (target[pos] == mapp[i + dir[k][0]][j + dir[k][1]])) {
                vis[i + dir[k][0]][j + dir[k][1]] ^= 1;
                search(pos + 1, i + dir[k][0], j + dir[k][1]);
                vis[i + dir[k][0]][j + dir[k][1]] ^= 1;
            }
        }
    }
private:
    int dir[4][2];
    int w, h;
    short **vis;
    char *target;
    int t_length;
    char **mapp;
    bool ans;
};