• leetcode------Word Search


    标题: Word Search
    通过率: 20.0%
    难度: 中等

    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.

    需要尝试从每一个位置起开始遍历,每次向四个方向开始遍历,如果四个方向都不满足时要将走过的路径回复,边界问题看代码:

    用一个index记录查了多少个了。

    迭代算法依然没有知道突破口:

     1 public class Solution {
     2     private int m,n;
     3     public boolean exist(char[][] board, String word) {
     4         m=board.length;
     5         n=board[0].length;
     6         boolean [][] visted=new boolean[m][n];
     7         for(int i=0;i<m;i++)
     8         for(int j=0;j<n;j++){
     9             if(dfs(board,word,0,i,j,visted))
    10                 return true;
    11         }
    12         return false;
    13     }
    14     public boolean dfs(char[][] board, String word,int index,int startx,int starty,boolean [][] visted){
    15         if(index==word.length())return true;
    16         if(startx<0||startx>=m||starty<0||starty>=n)return false;
    17         if(visted[startx][starty])return false;
    18         if(board[startx][starty]!=word.charAt(index))return false;
    19         visted[startx][starty]=true;
    20         boolean res=dfs(board,word,index+1,startx+1,starty,visted)||dfs(board,word,index+1,startx-1,starty,visted)||dfs(board,word,index+1,startx,starty+1,visted)||dfs(board,word,index+1,startx,starty-1,visted);
    21         visted[startx][starty]=false;
    22         return res;
    23     }
    24 }
  • 相关阅读:
    归并排序的java实现
    Hanoi问题java解法
    j2ee之Filter使用实例(页面跳转)
    java工具类之Graphics
    java程序设计之循环链表
    Java程序设计求出岁数
    Java程序设计之链表结构
    CENTOS 6 通过YUM升级GCC到4.7/4.8
    Object c的NSString的使用,创建,拼接和分隔,子string,substring
    Let’s Encrypt 最近很火的免费SSL 使用教程
  • 原文地址:https://www.cnblogs.com/pkuYang/p/4349936.html
Copyright © 2020-2023  润新知