[leetcode] 442. Find All Duplicates in an Array

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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?

样例：

Input:

[4,3,2,7,8,2,3,1]

Output:

[2,3]

在数组中找出所有重复出现的数，要求时间复杂度O(n),空间复杂度O(1)

直接贴支持率最高的代码

public class Solution {
    // when find a number i, flip the number at position i-1 to negative. 
    // if the number at position i-1 is already negative, i is the number that occurs twice.

    public 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;
    }
}

我的代码： 

利用数组输入的特点 1<=a(n)<=n,则可以直接利用原数组当hash表用。因为原数组都是正数，标为负数表示出现过，就可以找出所有出现过的数。 之后再把数组还原。

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

        }
        return list;
    }
}