LeetCode OJ 442. Find All Duplicates in an Array

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Description

Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.

Find all the elements that appear twice in this array.

Could you do it without extra space and in O(n) runtime?

Example

Input:
[4,3,2,7,8,2,3,1]

Output:
[2,3]

解题思路

与题目 LeetCode OJ 448. Find All Numbers Disappeared in an Array类似。
不使用额外的存储空间，在原数组中进行修改。将下标为|nums[i]|-1的数取反，即nums[ |nums[i]| - 1] = -|nums[ |nums[i]| - 1]|，最后如果nums[|nums[i]|-1] 已经为负，则说明|nums[i]|已经出现过一次。eg.

array 4 3 2 7 8 2 3 1 index 0 1 2 3 4 5 6 7 find -4 -3 -2 -7 8 2 -3 -1

当到第二个2（上表find斜黑体2）的时候，我们判断到下标为（2-1）的数nums[2-1] = -3，说明2这个数已经出现过了，所以是出现了两次。
将数组遍历了一遍，所以时间复杂度为O(n)。

代码

个人github代码链接

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