zoukankan      html  css  js  c++  java
  • Leetcode-Word Search

    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:

    Typical DFS problem. Need to consider the step of backtracking.

    Solution:

     1 public class Solution {
     2     public boolean exist(char[][] board, String word) {
     3         int xLen = board.length;
     4         if (xLen==0) return false;
     5         int yLen = board[0].length;
     6         if (yLen==0) return false;
     7 
     8         boolean[][] visited = new boolean[xLen][yLen];
     9         for (int i=0;i<xLen;i++)
    10             for (int j=0;j<yLen;j++)
    11                 visited[i][j] = false;
    12  
    13         char target = word.charAt(0);
    14         for (int i=0;i<xLen;i++)
    15             for (int j=0;j<yLen;j++)
    16                 if (board[i][j]==target){
    17                     //NOTE:We should consider about the case that the word.length is 1.
    18                     if (word.length()==1) return true;
    19                     visited[i][j]=true;
    20                     boolean find = searchRecur(board,visited,i,j,1,word);
    21                     visited[i][j]=false;
    22                     if (find) return true;
    23                 }
    24         return false;
    25     }
    26 
    27     public boolean searchRecur(char[][] board, boolean[][] visited, int x, int y, int cur, String word){
    28         int xLen = board.length;
    29         int yLen = board[0].length;
    30         char target = word.charAt(cur);
    31         if (cur==word.length()-1){
    32             if (x+1<xLen && board[x+1][y]==target && !visited[x+1][y]) return true;
    33             if (y+1<yLen && board[x][y+1]==target && !visited[x][y+1]) return true;
    34             if (x-1>=0 && board[x-1][y]==target && !visited[x-1][y]) return true;
    35             if (y-1>=0 && board[x][y-1]==target && !visited[x][y-1]) return true;
    36             return false;
    37         }
    38 
    39 
    40         if (x+1<xLen && board[x+1][y]==target && !visited[x+1][y]){
    41             visited[x+1][y] = true;
    42             boolean next = searchRecur(board,visited,x+1,y,cur+1,word);
    43             visited[x+1][y] = false;
    44             if (next) return true;
    45         }
    46 
    47         if (y+1<yLen && board[x][y+1]==target && !visited[x][y+1]){
    48             visited[x][y+1] = true;
    49             boolean next = searchRecur(board,visited,x,y+1,cur+1,word);
    50             visited[x][y+1] = false;
    51             if (next) return true;
    52         }
    53 
    54         if (x-1>=0 && board[x-1][y]==target && !visited[x-1][y]){
    55             visited[x-1][y] = true;
    56             boolean next = searchRecur(board,visited,x-1,y,cur+1,word);
    57             visited[x-1][y] = false;
    58             if (next) return true;
    59         }
    60 
    61         if (y-1>=0 && board[x][y-1]==target && !visited[x][y-1]){
    62             visited[x][y-1] = true;
    63             boolean next = searchRecur(board,visited,x,y-1,cur+1,word);
    64             visited[x][y-1] = false;
    65             if (next) return true;
    66         }
    67   
    68         return false;
    69     }
    70 }
  • 相关阅读:
    剑指offer二十九---最小的k个数
    Select2插件 点击、选中事件 解读
    Datatable插件的简单的使用方式 和 学习方式
    java map获取值方式
    mysql delete语句使用别名报错
    springmvc 添加@ResponseBody
    maven 创建后报错
    nodejs
    gulp
    Nodejs-express 4.0框架 简单介绍
  • 原文地址:https://www.cnblogs.com/lishiblog/p/4101012.html
Copyright © 2011-2022 走看看