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442. Find All Duplicates in an Array

时间:2018-02-05 19:40:32      阅读:222      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]

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

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

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