算法设计与应用基础：第一周（2）

442. Find All Duplicates in an Array

Description Submission Solutions

• Total Accepted: 17019
• Total Submissions: 32979
• Difficulty: Medium
• Contributors: shen5630

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?

Codes：

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

解题思路：关键点是nums中元素大小区间为[1,n]，n为nums元素个数，利用该特性，遍历nums，对于每一个元素，将其绝对值减1作为下标若对应元素为正数则说明该下标（也就是遍历的元素）第一次出现（此时将其修改为对应的相反数）；若为负数，则说明对于下标（遍历的元素）第二次出现，则将该下标加入答案数组。