Find All Numbers Disappeared in an Array问题及解法

问题描述：

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] inclusive that do not appear in this array.

Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.

示例：

Input:[4,3,2,7,8,2,3,1]Output:[5,6]

问题分析：

首先明确一点，对于一个[1,n]的数组，每一个索引都能找到1到n的对应值，若是缺一个值，那为该值的索引就不能通过num[i]取到，所以求解步骤分为以下两步：

1.先求解哪些nums的索引不能通nums[i]取到（迭代求解nums[i]映射，将能映射到的值设置为负的值）

2.将不能取到的索引值添加进结果集

过程详见代码：

class Solution {public:    vector<int> findDisappearedNumbers(vector<int>& nums) {        int len = nums.size();        for(int i = 0; i < len; i++)        {        int m = abs(nums[i])-1;            nums[m] = nums[m]>0 ? -nums[m] : nums[m];}vector<int> res;        for(int i = 0; i < len; i++) {            if(nums[i] > 0) res.push_back(i+1);        }        return res;    }};