697. Degree of an Array ----E

Q:

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: nums = [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: nums = [1,2,2,3,1,4,2]
Output: 6
Explanation: 
The degree is 3 because the element 2 is repeated 3 times.
So [2,2,3,1,4,2] is the shortest subarray, therefore returning 6.

 

Constraints:

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

A:

用一个map 来计算出degree来， 再用左右 index runner来跟着计算刚达到的degree

class Solution {
public:
    int findShortestSubArray(vector<int>& nums) {
        int degree = 0;
        unordered_map<int,int> map; // count of 
        for(auto k : nums){
            map[k] +=1;
            degree = max(degree, map[k]);
        }
        
        map.clear();
        int res = nums.size();
        int begin = 0;
        for(int end = 0;end<nums.size(); ++end)
        {
            int k = nums[end];
            map[k] += 1;
            if(map[k] == degree)
            {
                //loop till not equal
                while(nums[begin] != k)
                {
                    map[nums[begin]] -= 1;
                    ++begin;
                }
                res = min(res, end-begin+1);
                ++begin;
                map[k] -= 1;
            }
        }
        return res;
    }
};