leetcode刷題系列 448. Find All Numbers Disappeared in an Array

Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.

Find all the elements of [1, n] inclusive that do not appear in this array.

Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.

Example:

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

Output:
[5,6]

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

不停的做交換操作 直到每個位置跟該位置處的元素值一致爲止。