這兩道題很有意思，由於元素為1~n，因此每個元素的值-1（映射到0~n-1）就可以直接當做下標。這樣將 nums 中對應下標的元素 *-1 以i表示 index+1 這個元素出現過了，能節省存儲的空間。

448. Find All Numbers Disappeared in an Array

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

442. Find All Duplicates in an Array

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

LeetCode 448. Find All Numbers Disappeared in an Array & 442. Find All Duplicates in an Array