Given an array of integers, find out whether there are two distinct indices i and j in the array such that the difference between nums[i] and nums[j] is at most t and the difference between i and j is at most k. 这道题跟之前两道Contains Duplicate 包含重复值和Conta
Given an array of integers, find if the array contains any duplicates. Your function should return true if any value appears at least twice in the array, and it should return false if every element is distinct. 这道题不算难题,就是使用一个哈希表,遍历整个数组,如果哈希表里存在,返回fal
验证JS中是否包含重复元素,有重复返回true:否则返回false 方案一. function isRepeat(data) { var hash = {}; for (var i in data) { if (hash[data[i]]) { return true; } // 不存在该元素,则赋值为true,可以赋任意值,相应的修改if判断条件即可 hash[data[i]] = true; } return false; } 方案二. function isRepeat(arrs) { i
在上篇ORACLE查看表空间对象中,我介绍了如何查询一个表空间有那些数据库对象,那么我们是否可以查看某个数据文件包含那些数据库对象呢?如下所示 SELECT E.SEGMENT_TYPE AS SEGMENT_TYPE , E.SEGMENT_NAME AS SEGMENT_NAME , F.FILE_NAME AS FILE_NAME , SUM(E.BYTES)/1024/1024 AS SE
/*** * 去除List<PartsInfoDTO>列表中的重复对象 ~!! * @param list * @return */ public static List<PartsInfoDTO> removeDuplicate(List<PartsInfoDTO> list) { // Set<PartsInfoDTO> set = new HashSet<PartsInfoDTO>(); List<PartsInfoDTO> n
Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the absolute difference between i and j is at most k. 题目标签:Array, Hash Table 题目给了我们一个nums array 和一个 k, 让
下面小编就为大家带来一篇list集合去除重复对象的实现.小编觉得挺不错的,现在就分享给大家,也给大家做个参考.一起跟随小编过来看看吧 对象重复是指对象里面的变量的值都相等,并不定是地址.list集合存储的类型是基础类型还比较好办,直接把list集合转换成set集合就会自动去除. 当set集合存储的是对象类型时,需要在对象的实体类里面重写public boolean equals(Object obj) {} 和 public int hashCode() {} 两个方法. public Stri
原文链接:http://www.cnblogs.com/hanxi/archive/2012/07/25/2608068.html 前几天很不爽,因为C++中两个类中互相包含对方对象的指针编译时提示某一个类未定义...所以我就想啊想,这样也对,我的头文件都有#ifndef的,包含了一次就不能再包含了,以为就实现不了这样的功能,于是就改了设计方案: class A { public: A(B* pB):m_pB(pB) { } private: B* m_pB; }; class B { publ
Given an array of integers, find if the array contains any duplicates. Your function should return true if any value appears at least twice in the array, and it should return false if every element is distinct. Example 1: Input: [1,2,3,1] Output: tru
Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the absolute difference between i and j is at most k. Example 1: Input: nums = [1,2,3,1], k = 3 Output:
Given an array of integers, find out whether there are two distinct indices i and j in the array such that the absolute difference between nums[i] and nums[j] is at most t and the absolute difference between i and j is at most k. Example 1: Input: nu
TreeSet有序子类; HashSet无序子类 重复重复元素,Object对象是通过equals和hashCode来进行过滤的. 如果将上一篇提到中的例子中的TreeSet,换成HashSet,那么代码就不会过滤从的对象 HashSet无序子类,如何过滤重复的对象呢?(单个元素是可以过滤,但相同的对象不会) Person: public class Person implements Comparable<Person> { private String name; private int