• 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:
    ubuntu16.04下vim安装失败
    Sql Server函数全解(三)数据类型转换函数和文本图像函数
    Sql Server函数全解(二)数学函数
    Sql server 2008 中varbinary查询
    处理乱码问题
    快速排序
    《Java编程思想》笔记 第二章 一切都是对象
    1021: 组合数末尾的零
    11462
    The Bus Driver Problem
  • 原文地址:https://www.cnblogs.com/lishiblog/p/4101012.html
Copyright © 2020-2023  润新知