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?…

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?…

public class Solution { public IList<int> FindDisappearedNumbers(int[] nums) { Dictionary<int, int> dic = new Dictionary<int, int>(); ; i <= nums.Length; i++) { dic.Add(i, ); } ; i < nums.Length; i++) { dic[nums[i]]++; } var list =…

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?…

一个长度为N的数组,其中元素取值为1-N,求这个数组中没有出现的.1-N之间的数字. 要求无额外空间,O(n)时间复杂度. nums[i]=-1表示i数字已经出现过了 class Solution(object): def findDisappearedNumbers(self, nums): """ :type nums: List[int] :rtype: List[int] """ i=0 while i<len(nums): if…

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] class Solution { public: vector<int> findDisappearedNumbers(vec…

1.[LeetCode448]:448. 找到所有数组中消失的数字 题目分析: 1-n之间有重复的,有没出现的,有出现一次.使用hashmap,空间复杂度为O(n) 方法一:哈希表,但是空间复杂度超过了O(n) 思想: 可以用hashmap存储数据,建立映射. 从1-n去遍历,看hashmap中有没有出现元素. 有出现,下一个继续遍历,没有传入数组. 方法二: 使用数组本身,这样空间就不会多用 思想: 已知元素的范围是1-n,数组下标的范围是0-n-1.从头开始遍历,比如说4,把它转换为对应的下…