[leetcode] 442. Find All Duplicates in an Array

Question:

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]

Solution:

思路类似于翻牌子，充分利用条件 1 ≤ a[i] ≤ n，即数组元素的范围-1不会超出下标的范围，那么遇到遇到一个数，就把数字对应下标的那个牌子翻过来（取反），如果已经翻了过来，则说明此前已经遇到了这个数字，这个数字就是重复出现的那个。

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