原题
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.
思路：
遍历的过程中，存储我们想要的信息：
该元素是否出现的次数最多，在出现最多的前提下，它的最左最右出现的index差是否最小。
用一个dict数据结构来存储这些信息。
代码是：
class Solution:
def findShortestSubArray(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
dicts = {}
degreeEle = -1
degree = 0
minDiff = 0
for i in range(len(nums)):
if nums[i] not in dicts:
dicts[nums[i]] = [1,i,i]
#key-value : 元素 ，[次数,最左index,最右index]
else :
dicts[nums[i]][0] += 1
dicts[nums[i]][2] = i
if dicts[nums[i]][0] > degree: #在遍历中更新最值
degree = dicts[nums[i]][0]
degreeEle = nums[i]
minDiff = dicts[nums[i]][2] - dicts[nums[i]][1] + 1
elif dicts[nums[i]][0] == degree and dicts[nums[i]][2] - dicts[nums[i]][1] + 1 < minDiff:
degree = dicts[nums[i]][0]
degreeEle = nums[i]
minDiff = dicts[nums[i]][2] - dicts[nums[i]][1] + 1
print(degreeEle)
return minDiff
Python dict类型的基础：
