码迷,mamicode.com
首页 > 其他好文 > 详细

442. Find All Duplicates in an Array

时间:2018-02-05 19:59:27      阅读:135      评论:0      收藏:0      [点我收藏+]

标签:others   source   you   extra   turn   elements   integer   return   code   

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]

我的答案:

    public List<Integer> findDuplicates(int[] nums) {
        sort(nums);
        List<Integer> list = new ArrayList<>();
        for (int i = 1; i < nums.length; i++) {
            if(nums[i] == nums[i-1]){
                list.add(nums[i]);
            }
        }
        return list;
    }

大神答案:

    public static List<Integer> findDuplicates(int[] nums) {
        List<Integer> res = new ArrayList<>();
        for (int i = 0; i < nums.length; ++i) {
            int index = Math.abs(nums[i])-1;
            if (nums[index] < 0)
                res.add(Math.abs(index+1));
            nums[index] = -nums[index];
        }
        return res;
    }   

442. Find All Duplicates in an Array

标签:others   source   you   extra   turn   elements   integer   return   code   

原文地址:https://www.cnblogs.com/luozhiyun/p/8418844.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!