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.

class Solution(object):

    def findShortestSubArray(self, nums):

        """

        :type nums: List[int]

        :rtype: int

        """

        l, r = {}, {}  # 记录数字最左边和最右边的索引

        d = {}        #  记录每个数字的数量

        for i, num in enumerate(nums):

             if num not in l:

                 l[num] = i

              r[num] = i

              d[num] = d.get(num, 0) + 1

        degree = max(d.values())    # 找出值最大的数量

        res = len(nums)

        for i in d:

              if  d[i] == degree:

                   res = min(res, r[i] - l[i] + 1)

        return res