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]

Java Solution:

Runtime beats 85.92%

完成日期：09/19/2017

關鍵詞：Array

關鍵點：把num 和 num[num - 1] 做1對1的映射

 1 class Solution 
 2 {
 3     public List<Integer> findDuplicates(int[] nums) 
 4     {
 5         List<Integer> duplicates = new ArrayList<>();
 6         
 7         for(int num: nums)
 8         {
 9             int
 
 absNum = Math.abs(num);
10             
11             if(nums[absNum - 1] < 0) // if the number at position num - 1 is already negative    
12                 duplicates.add(absNum); // num is duplicate
13             else
14                 nums[absNum - 1] *= -1;
15         }
16         
17         return
 
 duplicates;
18     }
19 }

參考資料：

https://discuss.leetcode.com/topic/64735/java-simple-solution

LeetCode 題目列表 - LeetCode Questions List

LeetCode 442. Find All Duplicates in an Array （在數組中找到所有的重復項）