ArrayList类是List类下一种常用的子类,如果要判断容器里面的对象是否有相等,有两种方法. 下面是自定义的一个Student类,假设容器里重复是按照对象的两个属性都相等. /** * @author Wangchengan * */ public class Student { private String name; private int age; public String getName() { return name; } public void setName(String

［抄题］: Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements that appear twice in this array. Could you do it without extra space and in O(n) runtime? Example: Input: [4,

Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements that appear twice in this array. Could you do it without extra space and in O(n) runtime? Example: Input: [4,3,2,7,

Follow up for "Remove Duplicates":What if duplicates are allowed at most twice? For example,Given sorted array A = [1,1,1,2,2,3], Your function should return length = 5, and A is now [1,1,2,2,3]. 这道题是之前那道Remove Duplicates from Sorted Array 有序数组中

题目: 删除排序数组中的重复数字 给定一个排序数组,在原数组中删除重复出现的数字,使得每个元素只出现一次,并且返回新的数组的长度. 不要使用额外的数组空间,必须在原地没有额外空间的条件下完成.  样例 给出数组A =[1,1,2],你的函数应该返回长度2,此时A=[1,2]. 解题: 用Python直接搞 Python程序: class Solution: """ @param A: a list of integers @return an integer "&q

1.1 环境以及变量的定义.赋值.展开.删除 export:将一个变量导入到环境中:export PATH=$PATH:/home. readonly 讲一个变量设置为只读模式,在shell脚本中定义字面值常量特别有用:readonly days_per_week=7, readonly -p 打印当前环境只读模式变量. env可以对命令执行时的环境做更细致的控制:env -i PATH=$PATH HOME=$HOME awk '...' file1, 执行awk命令时的环境仅仅为env指定的

https://leetcode.com/problems/find-all-duplicates-in-an-array/description/ 参考:http://www.cnblogs.com/grandyang/p/4843654.html Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all th

//使用for循环 判断是否有重名 var len=$("li").length;//获取页面中所有li的数量 for(var i=0; i<len; i++){ oldname_wenjiaming=$("li:eq("+i+") #wenjianming").val(); //循环获得每个li下面的#wenjianming的值 if(checkname==oldname_wenjiaming){ alert("文件名有重复!&

[LeetCode] Remove Duplicates from Sorted Array 有序数组中去除重复项 描述 Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length. Do not allocate extra space for another array, you must do this

一.判断单个列表中的元素是否存在重复 使用set方法去重后,和原list进行对比,如果相等,那么说明原列表无重复,如果存在重复,说明列表存在重复 def is_repect_all(L): repeatList = []; setList = set(L); flag=True; if len(L) != len(setList): flag=False; print('列表的id存在重复,其中重复项及重复次数如下:'); for each_item in setList: re_count =

  Given a sorted array, remove the duplicates in-place such that each element appear only once and return the new length. Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory. Ex

Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length. Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length. Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra mem

Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length. Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra mem

public static int repeatedStringMatch(String A, String B) { //判断字符串a重复几次可以包含另外一个字符串b,就是不断叠加字符串a直到长度大于等于b,叠加一次计数+1 //看现在的字符串是否包含b,包含就返回,不会包含就再叠加一次,因为可能有半截的在后边,再判断,再没有就返回-1 int count = 0; StringBuilder sb = new StringBuilder(); while (sb.length() < B.l

知识点: .sort()方法用于对数组元素排序,并返回数组. var _arr = ['旅行箱', '旅行箱', '小米', '大米']; var _res = []; // _arr.sort(); for (var i = 0; i < _arr.length;) { var count = 0; for (var j = i; j < _arr.length; j++) { if (_arr[i] == _arr[j]) { count++; } } _res.push([_arr[i]

给定一个有序数组,你需要原地删除其中的重复内容,使每个元素只出现一次,并返回新的长度.不要另外定义一个数组,您必须通过用 O(1) 额外内存原地修改输入的数组来做到这一点.示例:给定数组: nums = [1,1,2],你的函数应该返回新长度 2, 并且原数组nums的前两个元素必须是1和2不需要理会新的数组长度后面的元素详见:https://leetcode.com/problems/remove-duplicates-from-sorted-array/description/ Java实现

这道题是LeetCode里的第26道题. 题目描述: 给定一个排序数组,你需要在原地删除重复出现的元素,使得每个元素只出现一次,返回移除后数组的新长度. 不要使用额外的数组空间,你必须在原地修改输入数组并在使用 O(1) 额外空间的条件下完成. 示例 1: 给定数组 nums = [1,1,2], 函数应该返回新的长度 2, 并且原数组 nums 的前两个元素被修改为 你不需要考虑数组中超出新长度后面的元素. 示例 2: 给定 nums = [0,0,,1,1,2,2,3,3,4], 函数应该返

一.题目 Description Given a sorted array, remove the duplicates in-place such that each element appear only once and return the new length. Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) e

和第一题不同的地方是,容忍两次重复 虽然题目上说只需要长度,但是否检测的时候如果数组不跟着改变也是不行的 没说清楚题意 自己是用双指针做的,看了大神的答案更简单 public int removeDuplicates(int[] nums) { int i = 0; for (int n : nums) if (i < 2 || n > nums[i-2]) nums[i++] = n; return i; }

Set<User> userSet = new HashSet<>(); User user1= new User("aa","11"); User user2= new User("aa","11"); userSet.add(user1); userSet.add(user2); System.out.println("set size"+userSet.size()); Ite