寻找数组中多余元素

题目来源:

点击打开链接

问题描述:

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]

我的解决方案:

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

思考:

此题与上一题寻找缺失元素如出一辙,所以还是选择用数组下标来对应表征元素的做法,不同的是由于本次需要获得多余的元素,他们相比其他元素而言是出现了两次,所以采用取负的方式可以得到,每当一个元素是两次取负变成正的时候,说明出现了两次,所以放入返回结果的容器中.时间复杂度位O(n).在阅读别人的解决方案的时候看到这样一种解法,虽然有两次循环,感觉不如我的方法快,但是比较好想好理解:(摘抄自leetcode solution,作者不详)

Firstly, we put each element x in nums[x - 1]. Since x ranges from 1 to N, then x - 1 ranges from 0 to N - 1, it won't exceed the bound of the array.
Secondly, we check through the array. If a number x doesn't present in nums[x - 1], then x is absent.

class Solution {
public:
    vector<int> findDuplicates(vector<int>& nums) {
        vector<int> res;
        int i = 0;
        while (i < nums.size()) {
            if (nums[i] != nums[nums[i]-1]) swap(nums[i], nums[nums[i]-1]);
            else i++;
        }
        for (i = 0; i < nums.size(); i++) {
            if (nums[i] != i + 1) res.push_back(nums[i]);
        }
        return res;
    }
};