题目描述：

LeetCode 697. Degree of an Array

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.

题目大意：

给定非空非负整数数组，数组的度是指元素的最大出现次数。

寻找最大连续区间，使得区间的度与原数组的度相同。

解题思路：

数组maxs记录元素的最大下标

数组mins记录元素的最小下标

数组cnts记录元素的出现个数

O(n)遍历即可

Python代码：

class Solution(object):
    def findShortestSubArray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        mins = {}
        maxs = {}
        cnts = collections.defaultdict(int)
        for idx, num in enumerate(nums):
            maxs[num] = max(maxs.get(num, -1), idx)
            mins[num] = min(mins.get(num, 0x7FFFFFFF), idx)
            cnts[num] += 1
        degree = max(cnts.values())
        ans = len(nums)
        for num in set(nums):
            if cnts[num] == degree:
                ans = min(ans, maxs[num] - mins[num] + 1)
        return ans

 

本文链接：http://bookshadow.com/weblog/2017/10/15/leetcode-degree-of-an-array/
请尊重作者的劳动成果，转载请注明出处！书影博客保留对文章的所有权利。

如果您喜欢这篇博文，欢迎您捐赠书影博客： ，查看支付宝二维码