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摘要: 题意 Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array w 阅读全文
posted @ 2016-09-09 17:05 Decouple 阅读(212) 评论(0) 推荐(0) 编辑
摘要: 题意 Given a digit string, return all possible letter combinations that the number could represent. A mapping of digit to letters (just like on the tele 阅读全文
posted @ 2016-09-09 16:43 Decouple 阅读(253) 评论(0) 推荐(0) 编辑
摘要: 题意 Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three intege 阅读全文
posted @ 2016-09-09 15:50 Decouple 阅读(204) 评论(0) 推荐(0) 编辑
摘要: 题意 Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum 阅读全文
posted @ 2016-09-09 15:31 Decouple 阅读(358) 评论(0) 推荐(0) 编辑
摘要: 题意 Given n non negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two en 阅读全文
posted @ 2016-05-11 20:25 Decouple 阅读(240) 评论(0) 推荐(0) 编辑
摘要: 详细介绍: https://tp iiita.quora.com/The Two Pointer Algorithm 阅读全文
posted @ 2016-05-11 20:15 Decouple 阅读(437) 评论(0) 推荐(0) 编辑
摘要: 题意 Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters 阅读全文
posted @ 2016-03-30 18:03 Decouple 阅读(256) 评论(0) 推荐(0) 编辑
摘要: 1 // pos = 当前处理的位置(一般从高位到低位) 2 // pre = 上一个位的数字(更高的那一位) 3 // status = 要达到的状态,如果为1则可以认为找到了答案,到时候用来返回, 4 // 给计数器+1。 5 // limit = 是否受限,也即当前处理这位能否随便取值。如56 阅读全文
posted @ 2015-09-15 12:43 Decouple 阅读(8894) 评论(1) 推荐(3) 编辑
摘要: #define MAXN 9999 #define MAXSIZE 10 #define DLEN 4 class BigNum { public: int a[500]; //可以控制大数的位数 int len; //大数长度 BigNum(){ len = 1;memset(a,0,sizeof 阅读全文
posted @ 2015-09-12 18:10 Decouple 阅读(410) 评论(0) 推荐(0) 编辑
摘要: 不要62 Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status Practice HDU 2089 Appoint description: System Crawler (2015-09 阅读全文
posted @ 2015-09-11 21:35 Decouple 阅读(252) 评论(0) 推荐(0) 编辑
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