1.LinkedHashSet 的概述和使用 llinkedHashSet 的特点: 是唯一能保证怎么存就怎么输出的 set 集合,并且去重复 1 LinkedHashSet<String> linkSet = new LinkedHashSet<>(); 2 /* LinkedHashSet可以保证元素唯一,并且可以保证有序(怎么存,就怎么取) */ 3 linkSet.add("f"); 4 linkSet.add("a"); 5 l

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

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

LeetCode:Combinations这篇博客中给出了不包含重复元素求组合的5种解法.我们在这些解法的基础上修改以支持包含重复元素的情况.对于这种情况,首先肯定要对数组排序,以下不再强调 修改算法1:按照求包含重复元素集合子集的方法LeetCode:Subsets II算法1的解释,我们知道:若当前处理的元素如果在前面出现过m次,那么只有当前组合中包含m个该元素时,才把当前元素加入组合 class Solution { public: void combine(vector<int> &a

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

Given an array of integers and an integer k, return true if and only if there are two distinct indices i and j in the array such that nums[i] = nums[j] and the difference between i and j is at most k. (Old Version) Given an array of integers and an i

Solution 动态规划. f(i)表示包含第i个字符的最长子串. 如果第i个字符没在之前出现过,那么f(i) = f(i - 1) + 1 如果第i个字符在之前出现过,这个时候应该分两种情况,假设与之前出现的距离为d,如果d>f(i - 1) 说明之前出现的字符不在f(i - 1)包含的字符串内,那么f(i) = f(i - 1) + 1, 如果d <= f(i - 1),说明包含的重复字符在f(i - 1)内,那么f(i) = d. Code class Solution { publi

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. 题目标签:Array, Hash Table 题目给了我们一个nums arr

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. 判断数组里面是否有重复的,很简单: class Solution { publ

首先将源码逐级找出来1.HashSet<String> hs=new HashSet<String>();         hs.add("hello");         hs.add("world");         hs.add("java");         hs.add("world");//因为是Set集合所以不带重复元素因为调用的是HashSet集合中的add方法,所以我们要找出来ad

package com.compare.test; import java.util.ArrayList;import java.util.Collections;import java.util.List; public class ListTest { public static List<Integer> createList1(){ List<Integer> list=new ArrayList<Integer>(); list.add(1); list.ad

public int findMin(int[] nums) { return findMin(nums, 0, nums.length - 1); } public int findMin(int[] nums, int left, int right) { int mid = (left + right)/2; if(left == right) { return nums[left]; } if(nums[left] < nums[right]) { return nums[left];

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. 这个类似前面两篇博客,只不过这里距离小于k的两个数的值小于t就可以满足要求,

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 difference between i and jis at most k. 这题上上一篇博客的延伸,问的是k长的距离内有没有两个数是相等的,类似一个滑动窗口问题,方法比较简单,使用一个map记下上

set函数去重 # -*- coding:utf-8 -*- srcTxt=open('1.txt','r').readlines() noRepeat=open('2.txt','w') st=set(srcTxt) noRepeat.write(''.join(st))

Python解法代码: class Solution: def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]: nums1.sort() nums2.sort() r = [] i = j = 0 while i < len(nums1) and j < len(nums2): if nums1[i] == nums2[j]: r.append(nums1[i]) i += 1 j += 1 elif n

给定两个数组,编写一个函数来计算它们的交集. 说明: 输出结果中每个元素出现的次数,应与元素在两个数组中出现的次数一致. 我们可以不考虑输出结果的顺序 def binarySearch(nums, target): ''' 在数组中二分查找指定元素 :param nums: :param target: :return: ''' left, right = 0, len(nums) - 1 while left <= right: mid = left + (right - left) // 2