LeetCode-Word Search

最新推荐文章于 2021-07-24 23:29:18 发布

转载最新推荐文章于 2021-07-24 23:29:18 发布 · 50 阅读

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原文链接：http://www.cnblogs.com/incrediblechangshuo/p/5931556.html

本文介绍了一个二维网格中查找指定单词的算法实现。该算法通过深度优先搜索(DFS)递归地检查每个单元格，判断是否能按顺序找到单词的所有字符，并确保同一字母单元格不被重复使用。

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 =

[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]
word = "ABCCED", -> returns true,
word = "SEE", -> returns true,
word = "ABCB", -> returns false.

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

转载于:https://www.cnblogs.com/incrediblechangshuo/p/5931556.html