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…
题目: 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) runt…
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?…
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?…
题目要求 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) run…
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] 题解 时间:O(n) 空间:O(1)(可以优化,不用 tem 的交换两个元素的值) class Solution { pub…
题目描述 给定n个数字的数组,里面的值都是1-n,但是有的出现了两遍,因此有的没有出现,求没有出现值这个数组中的值有哪些. 要求不能用额外的空间(除了返回列表之外),时间复杂度n 思路 因为不能用额外空间并且时间是O(n),所以不能用排序或者hash 通过在对应位置的值去确定下一个位置,一直到遍历完.其实自己也没有很成熟的思想. 比如对于[4,3,2,7,8,2,3,1]: A[0],4=>找到a[3],7=>a[6],3=>a[2]====注意有可能出现循环,行不通. 本来以为行不通,…
problem 448. Find All Numbers Disappeared in an Array solution: class Solution { public: vector<int> findDisappearedNumbers(vector<int>& nums) { vector<int> res; ; i<nums.size(); i++) { ; nums[tmp] = nums[tmp]> ? -nums[tmp] : n…
后面3个题都是限制在1-n的,所有可以不先排序,可以利用巧方法做.最后两个题几乎一模一样. 217. Contains Duplicate class Solution { public: bool containsDuplicate(vector<int>& nums) { int length = nums.size(); ) return false; sort(nums.begin(),nums.end()); ;i < length;i++){ ]) return tr…