【66】442. Find All Duplicates in an Array

转载于 2017-02-13 10:47:00 发布 · 78 阅读

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原文链接：http://www.cnblogs.com/93scarlett/p/6392844.html

文章标签：

#runtime

本文介绍了一种在不使用额外空间的情况下找到数组中所有重复出现的整数的方法，并提供了一个C++实现示例，该方法的时间复杂度为O(n)。

442. Find All Duplicates in an Array

Description Submission Solutions Add to List

Total Accepted: 15294
Total Submissions: 30028
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?

Example:

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

Output:
[2,3]

class Solution {
public:
//因为n[i]是在1到n之间，如果把n[i] - 1看成index的话，把其对i应的n[n[i]-1]换成负数，如果之前已经是负数，说明换过了（n[i] + 1重复）。
    vector<int> findDuplicates(vector<int>& nums) {
        vector<int> res;
        for(int i = 0; i < nums.size(); i++){
            int idx = abs(nums[i]) - 1;
            if(nums[idx] < 0){
                res.push_back(idx + 1);
            }
            nums[idx] = -nums[idx];
        }
        return res;
    }
};