let 79 Word search

最新推荐文章于 2025-11-05 18:25:18 发布

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本文介绍了一个二维网格中查找特定单词的问题解决方法。通过回溯法实现深度优先搜索，确保每个字母仅被使用一次。文章提供了完整的AC代码示例。

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.

主题思想： 读懂题意，每个字符只能适用一次，简单搜索题。搜索采用回溯法
AC 代码

class Solution {

    int [][] dir={{0,1},{0,-1},{1,0},{-1,0}};// right,left,down,up
    public boolean exist(char[][] board, String word) {
        if(board==null||board.length==0) return false;
        if(word==null||word.length()==0) return false;
        int m=board.length;
        int n=board[0].length;
        if(m*n<word.length())return false;


        boolean [][]mark=new boolean[m][n];

        boolean ans=false;
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(board[i][j]==word.charAt(0)){
                    mark[i][j]=true;
                    ans=dfs(board,mark,i,j,0,word);
                    mark[i][j]=false;
                    if(ans){
                        return ans;
                    }
                }
            }
        }
        return ans;
    }
    public boolean dfs(char[][] board,boolean[][] mark,int x,int y,int index,String word){

        if(index==word.length()-1){
            return true;
        }
        if(board[x][y]==word.charAt(index)){

            for(int i=0;i<4;i++){
                int nx=x+dir[i][0];
                int ny=y+dir[i][1];

                if(nx<0||nx>=board.length||ny<0||ny>=board[0].length||mark[nx][ny]) continue;

                if(board[nx][ny]==word.charAt(index+1)){
                    mark[nx][ny]=true;
                    if(dfs(board,mark,nx,ny,index+1,word)){
                        return true;
                    }
                    mark[nx][ny]=false;
                }
            }
        }
        return false;

    }
}