697. Degree of an Array

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.

 

数组里出现的频率最高的那个数，在子数组中也要出现那么多次，求子数组的最小长度

 

C++（52ms）：

class Solution {
public:
    int findShortestSubArray(vector<int>& nums) {
        int len = nums.size() ;
        if (len < 2)
            return len ;
        int Max = 0 ;
        unordered_map<int,int> startIndex , count ;
        int res = len ;
        for(int i = 0 ; i < len ; i++){
            if (startIndex.count(nums[i]) == 0)
                startIndex[nums[i]] = i ;
            count[nums[i]]++ ;
            if (count[nums[i]] > Max){
                res = i - startIndex[nums[i]] + 1 ;
                Max = count[nums[i]] ;
            }
            if (count[nums[i]] == Max){
                res = min(i - startIndex[nums[i]] + 1 , res) ;  
            }
        }
        return res ;
    }
};