題目：

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.

645. Set Mismatch

The set S originally contains numbers from 1 to n. But unfortunately, due to the data error, one of the numbers in the set got duplicated to another number in the set, which results in repetition of one number and loss of another number.

Given an array nums representing the data status of this set after the error. Your task is to firstly find the number occurs twice and then find the number that is missing. Return them in the form of an array.

思路：

448和645具有相似的思路。兩道題的共同點在於：元素的大小均為 [1,n]。448要求找出未出現的數字，645要求找出出現了兩次的數字和未出現的數字。

由於元素的下標為[0,n-1]，則元素的大小減去1得到的即為某個元素的下標，因此可以利用元素大小與下標之間的關系尋找特殊的數字。

對於448，遍歷整個數組，通過元素的大小減去1得到下標。若該下標對應的元素為正，則將其乘以-1，變為負數；若該下標對應的元素為負，證明該下標之前已經出現過了一次，不作處理。通過這一次的遍歷，仍然為正的元素所對應的下標再加1即為未出現過的元素。

對於645，遍歷整個數組，通過元素的大小減去1得到下標。若該下標對應的元素為正，則將其乘以-1，變為負數；若該下標對應的元素為負，證明該下標之前已經出現過了一次，將該下標+1加入result中。通過這一次的遍歷，仍然為正的元素所對應的下標再加1即為未出現過的元素。

代碼：

448.

 1 class Solution {
 2 public:
 3     vector<int> findDisappearedNumbers(vector<int>& nums) {
 4         vector<int> ans;
 5         for (int i = 0; i < (signed) nums.size(); i++) {
 6             int index = abs(nums[i]) - 1;
 7             if (nums[index] > 0)
 8                 nums[index] *= -1;
 9             else
10                 continue;
11         }
12         for (int i = 0; i < (signed) nums.size(); i++)
13             if (nums[i] > 0)
14                 ans.push_back(i + 1);
15         return ans;
16     }
17 };

645.

 1 class Solution {
 2 public:
 3     vector<int> findErrorNums(vector<int>& nums) {
 4         vector<int> ans;
 5         for (int i = 0; i < (signed) nums.size(); i++) {
 6             int index = abs(nums[i]) - 1;
 7             if (nums[index] > 0) {
 8                 nums[index] *= -1;
 9             } else {
10                 ans.push_back(index + 1);
11             }
12         }
13         for (int i = 0; i < (signed) nums.size(); i++) {
14             if (nums[i] > 0)
15                 ans.push_back(i + 1);
16         }
17         return ans;
18     }
19 };

448. Find All Numbers Disappeared in an Array&&645. Set Mismatch