LeetCode 442. Find All Duplicates in an Array

题目：
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]

思路：
具体思路见注释。

代码：

class Solution {public:    vector<int> findDuplicates(vector<int>& nums) {        vector<int>ret;        if(nums.size()<=1)//如果nums的长度小于等于1，直接返回空vector            return ret;        int i;        for(i=0;i<nums.size();i++){//所有的数减1            nums[i]-=1;        }        i=0;        while(i<nums.size())        {            if(nums[i]!=i&&nums[i]!=nums[nums[i]])//将所有数放到对应数值的索引位置                swap(nums[i],nums[nums[i]]);            else                i++;        }        for(i=0;i<nums.size();i++)        {            if(nums[i]!=i)//如果对应索引位置的数不与索引相等                ret.push_back(nums[i]+1);//将刚开始减去的1加上，并push_backt到ret里        }        return ret;    }};

**输出结果：**149ms