448.Find All Numbers Disappeared in an Array

Description：
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)

Input: [4,3,2,7,8,2,3,1]
Output: [5,6]

方法1：額外空間，不符合題目要求

• Time complexity : $O(n$
  ) O\left ( n \right )
• Space complexity : $O\left(n\right)$
  O\left ( n \right )
  思路：
  第一直覺就是建標誌陣列，出現過的標記為1，最後結果是非1的索引

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

方法2：最優解，符合題意

• Time complexity : $O\left ( n \right )$
• Space complexity : $O\left ( 1 \right )$
  思路：
  利用改變元素正負號和取絕對值的方法，取代新建標誌陣列

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