448. Find All Numbers Disappeared in an Array Input: [4,3,2,7,8,2,3,1] Output: [5,6] 思路:把数组的内容和index进行一一对应映射,映射规则是取反,由此可知,重复出现两次的数字会变为正,出现一次的为负. 需要注意的是,如果不能增加额外空间的话,要在本数组上面进行映射,这时候就需要内容-1=index,因为index是从0开始,内容从1开始 class Solution { public: vector<int>…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 公众号:负雪明烛 本文关键词:Leetcode, 力扣,476, 补数,二进制,Python, C++, Java 目录 题目描述 解题方法 方法一:取反 方法二:异或 方法三:二进制字符串 总结 日期 题目地址:https://leetcode.com/problems/number-complement/ Difficulty: Easy 题目描述 Given a positive…

Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements of [1, n] inclusive that do not appear in this array. Could you do it without extra space and in O(n) runtime?…

原题链接 https://leetcode.com/problems/number-complement/ 原题 Given a positive integer, output its complement number. The complement strategy is to flip the bits of its binary representation. Note: The given integer is guaranteed to fit within the range o…

题目链接: https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/description/ 题目描述: Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements of [1, n] incl…

problem 476. Number Complement solution1: class Solution { public: int findComplement(int num) { //正数的补数是对应二进制各位翻转,且从非零高位开始. bool start = false; ; i>=; i--)//err. { <<i)) start = true; <<i); } return num; } }; 参考 1. Leetcode_476. Number Com…

448. 找到所有数组中消失的数字 给定一个范围在 1 ≤ a[i] ≤ n ( n = 数组大小 ) 的 整型数组,数组中的元素一些出现了两次,另一些只出现一次. 找到所有在 [1, n] 范围之间没有出现在数组中的数字. 您能在不使用额外空间且时间复杂度为O(n)的情况下完成这个任务吗? 你可以假定返回的数组不算在额外空间内. 示例: 输入: [4,3,2,7,8,2,3,1] 输出: [5,6] 通过次数29,638提交次数51,956 class Solution { public Li…

题目 给定一个范围在  1 ≤ a[i] ≤ n ( n = 数组大小 ) 的 整型数组,数组中的元素一些出现了两次,另一些只出现一次. 找到所有在 [1, n] 范围之间没有出现在数组中的数字. 您能在不使用额外空间且时间复杂度为O(n)的情况下完成这个任务吗? 你可以假定返回的数组不算在额外空间内. 示例: 输入: [4,3,2,7,8,2,3,1] 输出: [5,6] 来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/find-all-…

My first reaction is to have an unlimited length of bit-array, to mark existence. But if no extra mem is allowed, we can simply use 'sign' on each index slot to mark the same info.. it is a common technique. class Solution { public: vector<int> find…

Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements of [1, n] inclusive that do not appear in this array. Could you do it without extra space and in O(n) runtime?…