[LeetCode124]Word Search

最新推荐文章于 2025-08-09 17:01:50 发布

原创最新推荐文章于 2025-08-09 17:01:50 发布 · 786 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #递归

Leetcode 同时被 2 个专栏收录

287 篇文章

订阅专栏

LeetCode Questions

178 篇文章

订阅专栏

本文介绍了一种在二维网格中查找指定单词的算法实现。通过递归搜索相邻单元格内的字母来构造目标单词，确保同一字母单元格不会被重复使用。文章提供了详细的逻辑流程与 Java 和 C++ 的代码示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

Analysis:

The idea of this question is as follows:
(1) Find the 1st element of the word in the board.
(2) For each position found where the 1st element lies, recursively do:
(i) Search the around cell to see if the next element exists. (4 directions: (i-1,j),(i+1,j),(i,j-1),(i,j+1) )
(ii) If the word ends, return true.
(3) Return false if no matching found.
Note: A visited matrix is needed to store the positions where have already been visited. Details can be found in code.

Java

public class Solution {
    int[][] visited;
	public boolean exist(char[][] board, String word) {
        if(word.length()==0) return false;
		if(board.length==0 || board[0].length==0) return false;
        int row = board.length;
        int col = board[0].length;
        visited = new int [row][col];
        for(int i=0;i<row;i++){
        	for(int j=0;j<col;j++){
        		if(board[i][j] == word.charAt(0)){
        			visited[i][j] = 1;
        			if(search(board, word, -1, i, j, 1))
        				return true;
        			visited[i][j] = 0;
        		}
        	}
        }
        
        return false;
    }
	//op 0123 up down left right
	public boolean search(char[][] board, String word,int op,int i, int j, int matchLen){
		if(matchLen == word.length()) return true;
		int row = board.length;
		int col = board[0].length;
		if(i+1<row && op!=0){//down
			if(visited[i+1][j]==0 && board[i+1][j]==word.charAt(matchLen)){
				visited[i+1][j] =1;
				if(search(board, word, 1, i+1, j, matchLen+1))
					return true;
				visited[i+1][j]=0;
			}
		}
		if(i-1>=0 && op!=1){//up
			if(visited[i-1][j]==0 && board[i-1][j] == word.charAt(matchLen)){
				visited[i-1][j] = 1;
				if(search(board, word, 0, i-1, j, matchLen+1))
					return true;
				visited[i-1][j] = 0;
			}
		}
		if(j+1<col && op!=2){
			if(visited[i][j+1] ==0 && board[i][j+1] == word.charAt(matchLen)){
				visited[i][j+1] = 1;
				if(search(board, word, 3, i, j+1, matchLen+1))
					return true;
				visited[i][j+1] = 0;
			}
		}
		if(j-1>=0 && op!=3){
			if(visited[i][j-1] ==0 && board[i][j-1] == word.charAt(matchLen)){
				visited[i][j-1] = 1;
				if(search(board, word, 2, i, j-1, matchLen+1))
					return true;
				visited[i][j-1] = 0;
			}
		}
		return false;
	}
}

c++

class Solution {
public:
    bool searchWord(vector<vector<char>> &board,
            string word,
            int i,
            int j,
            int matchLen,
            int op,// 0 up, 1 down, 2 left, 3 right
            vector<vector<int>> &visited){
    if(matchLen == word.size()) return true;
    int row = board.size();
    int col = board[0].size();
    if(i+1<row && op!=0){
        if(visited[i+1][j]==0 && board[i+1][j]==word[matchLen]){
            visited[i+1][j] = 1;
            if(searchWord(board,word,i+1,j,matchLen+1,1,visited))
                return true;
            visited[i+1][j] = 0;
        }
    }
    if(i-1>=0 && op!=1){
        if(visited[i-1][j]==0 && board[i-1][j]==word[matchLen]){
            visited[i-1][j] = 1;
            if(searchWord(board,word,i-1,j,matchLen+1,0,visited))
                return true;
            visited[i-1][j] = 0;
        }
    }
    if(j+1<col && op!=2){
        if(visited[i][j+1]==0 && board[i][j+1]==word[matchLen]){
            visited[i][j+1] = 1;
            if(searchWord(board,word,i,j+1,matchLen+1,3,visited))
                return true;
            visited[i][j+1] = 0;
        }
    }
    if(j-1>=0 && op!=3){
        if(visited[i][j-1]==0 && board[i][j-1]==word[matchLen]){
            visited[i][j-1] = 1;
            if(searchWord(board,word,i,j-1,matchLen+1,2,visited))
                return true;
            visited[i][j-1] = 0;
        }
    }
    return false;
}
bool exist(vector<vector<char> > &board, string word) {
    int row = board.size();
    int col = board[0].size();
    vector<vector<int>> visited(row,vector<int>(col,0));
    for(int i=0;i<row;i++){
        for(int j=0;j<col;j++){
            if(board[i][j] == word[0])
                if(word.size()==1) return true;
                else{
                    visited[i][j] = 1;
                    if(searchWord(board,word,i,j,1,-1,visited))
                        return true;
                    visited[i][j] = 0;
                }
        }
    }
    return false;
}
};