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1. 括号匹配 | |
给定一堆括号序列,请判断是不是合法的括号序列 | |
()合法 | |
(() 不合法 | |
(())()合法 | |
分析空间和时间 | |
2. | |
Scramble with Friends是一款老少咸宜的益智类游戏 (下载地址:Apple Store Google Play) | |
玩法是,给定一个4x4的方块矩阵,每个格子上都有一个字母。玩家需要来通过串连这些邻接(上,下,左,右,左上,左下,右上,右下)的方块组合成合法的单词(如上图的scrumble)。在规定时间内,找出的单词越多越好。(Please ignore the number on the tiles) | |
假如聪明的小伙伴你,是这个游戏的开发者。我们随机给你生成了一个nxn的字母矩阵,我们还给你一本单词本,里面列举了所有合法的英语词语(assume不多不少,)。你的任务是写一个程序找出这个矩阵可能出现的所有合法单词。 | |
java: | |
String[] solve(char[][] matrix, String[] DanCiBook){ | |
//you code starts here. | |
} | |
Time Limits : 1s (just kidding, i dont have OJ for this) | |
样例输入: | |
matrix: | |
[[o c], | |
[b a]] | |
DanCiBook: | |
[cao, condom, cab, sexy, apple, fuck, cnm, nmb, cnmb, renjianbuchai, xidapuben,xiao, kan, shi, ge, da, sha, bi] | |
样例输出: 按alphabetic排序 | |
[cab, cao] |
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//1 | |
public class Solution { | |
public boolean validateParenthesis(String p) { | |
if(p == null) return false; | |
int left = 0; | |
for(int i = 0; i < p.length(); i++){ | |
if(p.charAt(i) == '(') left++; | |
else if(p.charAt(i) == ')') left--; | |
if(left < 0) return false; | |
} | |
return left == 0; | |
} | |
} | |
//2 |
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