首页 > 编程语言 > 详细

442. 找出数组中重复的元素 Find All Duplicates in an ArrayGiven an array of integers

时间:2017-03-11 12:50:42      阅读:718      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]

题意:找出数组中重复的元素
关键:1 ≤ a[i] ≤ n  和 appear twice
解法:nums[Math.Abs(nums[i]) - 1] *= -1 如果 nums[Math.Abs(nums[i]) - 1] 小于0 则nums[i]重复
  1. public class Solution {
  2. public IList<int> FindDuplicates(int[] nums) {
  3. List<int> list = new List<int>();
  4. for (int i = 0; i < nums.Length; i++) {
  5. nums[Math.Abs(nums[i]) - 1] *= -1;
  6. if (nums[Math.Abs(nums[i]) - 1] > 0) {
  7. list.Add(Math.Abs(nums[i]));
  8. }
  9. }
  10. return list;
  11. }
  12. }





442. 找出数组中重复的元素 Find All Duplicates in an ArrayGiven an array of integers

原文:http://www.cnblogs.com/xiejunzhao/p/fd62258e250caf4f07840c19bb9fac0d.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!