697. Degree of an Array

题目

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.

思路

本题分三步来解决这个问题：
第一步是利用一个map来存储值与对应的索引，
第二步是扫描map找到这个数组的度
第三步是扫描数组，找到最短的子数组

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 maxfreq = 0;
        for(auto it=mp.begin();it!=mp.end();it++)
        {
            maxfreq = max(maxfreq,int(it->second.size()));
            //cout<<it->second.size()<<endl;
        }
        int minlen = nums.size();
        for(auto it=mp.begin();it!=mp.end();it++)
        {
            if(it->second.size()==maxfreq)
            {
                minlen = min(minlen,it->second.back()-it->second[0]+1);
            }
        }
        return minlen;
    }
};