【WordSearch】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.
public class Solution {
public boolean exist(char[][] board, String word) {
if(word.length()==0)
return true;
if(board.length==0)
return false;
boolean[][] bol = new boolean[board.length][board[0].length];
for(int i=0;i<board.length;i++){
for(int j=0;j<board[i].length;j++){
if(board[i][j]==word.charAt(0)&&findExist(i,j,0,board,bol,word)){
return true;
}
}
}
return false;
}
private boolean findExist(int x, int y,int wa, char[][] board, boolean[][] bol, String word) {
if(wa>=word.length())
return true;
if(x<0||y<0)
return false;
if(x>=board.length||y>=board[x].length)
return false;
if(bol[x][y])
return false;
if(board[x][y]!=word.charAt(wa))
return false;
bol[x][y]=true;
boolean re =findExist(x+1,y,wa+1,board,bol,word)||findExist(x,y+1,wa+1,board,bol,word)||
findExist(x-1,y,wa+1,board,bol,word)||findExist(x,y-1,wa+1,board,bol,word);
bol[x][y]=false;
return re;
}
}
【WordSearch】Word Search的更多相关文章
- 【leetcode】Word Search
Word Search Given a 2D board and a word, find if the word exists in the grid. The word can be constr ...
- 【leetcode】Word Search (middle)
今天开始,回溯法强化阶段. Given a 2D board and a word, find if the word exists in the grid. The word can be cons ...
- 【leetcode】Word Search II(hard)★
Given a 2D board and a list of words from the dictionary, find all words in the board. Each word mus ...
- 【Leetcode】【Medium】word search
Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from l ...
- 【数组】word search
题目: Given a 2D board and a word, find if the word exists in the grid. The word can be constructed fr ...
- 【HDU2222】Keywords Search AC自动机
[HDU2222]Keywords Search Problem Description In the modern time, Search engine came into the life of ...
- 【CF528D】Fuzzy Search(FFT)
[CF528D]Fuzzy Search(FFT) 题面 给定两个只含有\(A,T,G,C\)的\(DNA\)序列 定义一个字符\(c\)可以被匹配为:它对齐的字符,在距离\(K\)以内,存在一个字符 ...
- 【计算机视觉】Selective Search for Object Recognition论文阅读3
Selective Search for Object Recoginition surgewong@gmail.com http://blog.csdn.net/surgewong 在前 ...
- 【HDU2222】Keywords Search(AC自动机)
Problem Description In the modern time, Search engine came into the life of everybody like Google, B ...
随机推荐
- LeetCode OJ-- Simplify Path **
https://oj.leetcode.com/problems/simplify-path/ 对linux路径的规范化,属于字符串处理的题目.细节很多. #include <iostream& ...
- ros pub sub ("~")
base_velocity_smoother.cpp remap ("~") 订阅: "cmd_vel" ...
- ubuntu下打开windows里的txt文件乱码解决
是编码问题引起的问题: Linux下默认的编码是UTF-8,而Windows下默认的编码是GB2312/GBK.执行如下第一条语句即可 gsettings set org.gnome.gedit.pr ...
- android-samples-mvp
Model–view–presenter (MVP)介绍 mvp在wiki上的介绍为 Model 定义用户界面所需要被显示的数据模型,一个模型包含着相关的业务逻辑 View View不应该处理业务 ...
- Markdown的css样式源码
http://www.cnblogs.com/zhangjk1993/p/5442676.html https://github.com/zhangjikai/markdown-css https:/ ...
- docker 如何清理垃圾呢
应用容器在宿主机上长期运行,应用实例启停容器,会产生大量的停止的容器,无容器使用的数据卷.网络配置,无容器依赖的镜像,这些垃圾日积月累,会影响到宿主机的运行状态,造成机子卡顿等现象.因此,需要对这些宿 ...
- sh脚本——#!/bin/bash
#!/bin/bash是指此脚本使用/bin/bash来解释执行. 其中,#!是一个特殊的表示符,其后,跟着解释此脚本的shell路径. bash只是shell的一种,还有很多其它shell,如:sh ...
- flex操作XML,强力总结帖
初始化XML对象 XML对象可以代表一个XML元素.属性.注释.处理指令或文本元素.在ActionScript 3.0中我们可以直接将XML数据赋值给变量: var myXML:XML = &l ...
- POJ 2309 BST
BST Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 8565 Accepted: 5202 Description C ...
- 虚幻4Matinee功能 基本概念及简单演示样例(Sequence编辑器)
虚幻4提供的Matinee功能十分强大,能够用来制作动画.录制视频. 它的核心想法是在Matinee编辑器内提供一套自己的时间坐标系,在这个相对时间内通过调节actor的属性来改变actor的状态,进 ...