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.
出现次数最多元素的idx差值：
class Solution(object):
def findShortestSubArray(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
d=collections.defaultdict(list)
for i, v in enumerate(nums):
d[v].append(i)
length, m=0, 0
for l in d.values():
if len(l)==length:
m=min(m, l[-1]-l[0]+1)
elif len(l)>length:
length=len(l)
m=l[-1]-l[0]+1
return m