645. Set Mismatch-优快云博客

本文介绍了一种使用哈希表解决数组中重复和缺失数字问题的方法。在一个包含1到n的整数集合中，由于数据错误导致一个数字重复出现而另一个数字丢失。通过给出的错误数据状态数组，任务是找到重复的数字以及丢失的数字。

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.

Example 1:

Input: nums = [1,2,2,4]
Output: [2,3]

Note:

The given array size will in the range [2, 10000].
The given array's numbers won't have any order.

用了哈希表，其实在做的时候就应该意识到，可以排个序，最后肯定是nums[i]=i+1

class Solution {

public:
      vector<int> findErrorNums(vector<int>& nums) {
       vector<int> temp;
       unordered_map<int,int> hashmap;
       for(int i=0;i<nums.size();i++){
           if(hashmap.find(nums[i])==hashmap.end())
               hashmap.insert(pair<int,int>(nums[i],1));
           else{
               hashmap[nums[i]]++;
               if(hashmap[nums[i]]==2)
                   temp.push_back(nums[i]);
           }
       }
       for(int i=1;i<=nums.size();i++){
           if(hashmap.find(i)==hashmap.end())
               temp.push_back(i);
       }
       return temp;
    }
};