首页 > 其他 > 详细

[leetcode-442-Find All Duplicates in an Array]

时间:2017-05-12 20:55:53      阅读:258      评论:0      收藏:0      [点我收藏+]

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:

The concept here is to negate each number‘s index as the input is 1 <= a[i] <= n (n = size of array).

Once a value is negated, if it requires to be negated again then it is a duplicate.

vector<int> findDuplicates(vector<int>& nums)
     {
         vector<int>ret;
         for (int i = 0; i < nums.size();i++)
         {
             int index = abs(nums[i]) - 1;
             if (nums[index] > 0)
             {
                 nums[index] = - nums[index];
             }
             else
             {
                 ret.push_back(abs(nums[i]));
             }
         }
         return ret;
     }

参考:

https://discuss.leetcode.com/topic/64805/java-easy-to-understand-solution-without-extra-space-and-in-o-n-time

[leetcode-442-Find All Duplicates in an Array]

原文:http://www.cnblogs.com/hellowooorld/p/6847120.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!