697. Degree of an Array

算出degree相同的最小子列表.
degree是一个列表中出现相同元素的最大次数

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

从原列表的degree开始, 求出每一个子列表, 计算列表的degree, 如果两个degree相同, 就是所求的结果

class Solution(object):
    # 求出列表的degree
    def findDegree(self, nums):
        return  max([nums.count(i) for i in set(nums)])

    def findShortestSubArray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
       
        rsts = []
        degree = self.findDegree(nums)
        for n in range(degree, len(nums)+1):
            for k in range(n,len(nums)+1):
                sub_degress = self.findDegree(nums[k-n:k])
                #print(n, nums[k-n:k])
                if sub_degress == degree:
                    return n

时间复杂度 O(n3),超时

# 计算最大degree的元素的最大距离
def findShortestSubArray(self, nums):
        c = collections.Counter(nums)
        first, last = {}, {}
        for i, v in enumerate(nums):
            first.setdefault(v, i)
            last[v] = i
        degree = max(c.values())
        return min(last[v] - first[v] + 1 for v in c if c[v] == degree)