[LeetCode] 448. Find All Numbers Disappeared in an Array

题目链接: https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/

Description

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.

Example:

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

解题思路

遍历一遍数组，出现数字 - 1当做数组下标，将该下标在数组中的值存为负数，用来表示这个数字出现过。
例如，题目样例中的第一个数 4，就将数组下标为 4 - 1 = 3 的内容存为负数，即 nums[3] = nums[3] > 0 ? -nums[3] : nums[3]。然后再次遍历一遍数组，存储值为正数的数组下标 + 1，即为缺失的值。
还是拿题目样例来说明，按照上面的思路，要循环遍历两次。第一次遍历后数组内容变为:

[-4, -3, -2, -7, 8, 2, -3, -1]

第二次遍历，就找出了两个正数的下标，4 和 5，分别加 1，得缺失的值为 5 和 6。

Code

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