697. Degree of an Array

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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.

找一个最短的连续子数组，是的里面出现频率最多的数字的频率等于原数组出现频率最多的数字的频率。

on 用hashmap也就是python的dict来记录每个元素的出现次数以及第一次出现的位置和最后一次出现的位置，然后再维护一个频率最大值。

遍历每个频率等于最大频率的值，看看他最晚出现-最早出现+1长度是不是最短

class Solution(object):
    def findShortestSubArray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        count = {}
        first = {}
        last = {}
        max_count = 0
        for i in range(len(nums)):
            if nums[i] not in count:
                count[nums[i]] = 1
            else:
                count[nums[i]] += 1
            max_count = max(max_count, count[nums[i]])
            if nums[i] not in first:
                first[nums[i]] = i
            last[nums[i]] = i
        ans = len(nums)
        for key,value in count.items():
            if value == max_count:
                ans = min(ans, last[key] - first[key] + 1)
        return ans

697. Degree of an Array

标签：app   pos   task   lse   bec   ati   you   item   exp   

原文地址：https://www.cnblogs.com/whatyouthink/p/13232612.html