leetcode961. Find the elements repeated N times in the array with length 2n

Give you an array of integers nums, The array has the following properties ：

• nums.length == 2*n.
• nums contain n+1 A different element .
• nums There happens to be an element repetition in n Time .

Find and return duplicates n The next element .

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

Ideas ：
Finding the target element requires two iterations of the array ：

1. First traversal , Count the number of occurrences of each element .
2. Second traversal , Return to appear n Secondary elements , That is, the target element .

The code is as follows ：

class Solution {
    
public:
	int repeatedNTimes(vector<int>& nums) {
    
		unordered_map<int, int> countMap;
		//1、 First traversal , Count the number of times 
		for (auto e : nums)
		{
    
			countMap[e]++;
		}
		//2、 Second traversal , Return to appear n Secondary elements 
		for (auto e : countMap)
		{
    
			if (e.second == nums.size() / 2)
				return e.first;
		}
		return -1; // Make sure that the compilation passes （ All paths have return values ）
	}
};

Explain ： According to the question , When executing the code, you must find an occurrence n The second element is returned , This is called execution logic , But when the code is compiled , According to the compilation logic , We must make every logic have a return value , So you need to return an arbitrary value at the end of the code , Its purpose is to make the code compile smoothly , Although we know that the code will not run here to return .