697. Degree of an Array

题目描述：

Given a non-empty array of non-negative integers nums, the degree of this array is defined as the maximum frequency of any one of its elements.

Your task is to find the smallest possible length of a (contiguous) subarray of nums, that has the same degree as nums.

Example 1:
Input: [1, 2, 2, 3, 1]
Output: 2
Explanation:
The input array has a degree of 2 because both elements 1 and 2 appear twice.
Of the subarrays that have the same degree:
[1, 2, 2, 3, 1], [1, 2, 2, 3], [2, 2, 3, 1], [1, 2, 2], [2, 2, 3], [2, 2]
The shortest length is 2. So return 2.
Example 2:
Input: [1,2,2,3,1,4,2]
Output: 6
Note:

nums.length will be between 1 and 50,000.
nums[i] will be an integer between 0 and 49,999.

解题思路：

1. 要想找到出现次数最多的元素，必须把数组扫描一遍，且要记录下每个元素出现的次数，这里借用map实现最好。可以同时用vector记录下初始位置和结束位置，这样后面在计算长度时可以直接使用。

2. 遍历map得到最大数量的元素，计算长度返回最短长度。

解题代码：

class Solution {
public:
    int findShortestSubArray(vector<int>& nums) {
        unordered_map<int,vector<int>> mp;
        for(int i=0;i<nums.size();i++) mp[nums[i]].push_back(i);
        int degree=0;
        for(auto it=mp.begin();it!=mp.end();it++) degree=max(degree,int(it->second.size()));
        int shortest=nums.size();
        for(auto it=mp.begin();it!=mp.end();it++)
        {
            if(it->second.size()==degree)
            {
                shortest=min(shortest,it->second.back()-it->second[0]+1);
            }
        }
        return shortest;
    }
};