Word Search, Leetcode 解题笔记

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.

DFS变种题,直接看代码比较容易理解:

public class Solution {
    boolean[][] visited;
    public boolean exist(char[][] board, String word) {
        int m = board.length;
        int n = board[0].length;
        visited = new boolean[m][n];
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(dfs(board, word, i, j, 0)){
                    return true;
                }
            }
        }
        return false;
    }
    
    public boolean dfs(char[][] board, String word, int i, int j, int n){
        if(n == word.length()){
            return true;
        }
        if(i < 0 || i > board.length-1 || j < 0 || j > board[0].length-1){
            return false;
        }
        if(visited[i][j]){
            return false;
        }
        if(word.charAt(n) != board[i][j]){
            return false;
        }
        visited[i][j] = true;
        boolean result = dfs(board, word, i+1, j, n+1) || dfs(board, word, i-1, j, n+1) || dfs(board, word, i, j+1, n+1) || dfs(board, word, i, j-1, n+1);
        visited[i][j] = false;
        return result;
    }
}
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