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.

class Solution(object):
    def findShortestSubArray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        # use pos_map to record num's start pos and end pos in nums
        pos_map = {}
        for i,n in enumerate(nums):
            if n not in pos_map:
                pos_map[n] = [i, i]
            else:
                pos_map[n][1] = i
                
        # find max_freq number (maybe have many) 
        cnt = collections.Counter(nums)
        max_freq = 0
        max_freq_num = []
        for k,v in cnt.iteritems():
            if v > max_freq:
                max_freq = v
                max_freq_num = [k]                
            elif v == max_freq:
                max_freq_num.append(k)
                    
        # find min degree
        return min(pos_map[n][1]-pos_map[n][0]+1 for n in max_freq_num)                            

精简代码：

class Solution:
    def findShortestSubArray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        pos_map = {}
        for i,n in enumerate(nums):
            if n not in pos_map:
                pos_map[n] = [i, i]
            else:
                pos_map[n][1] = i            
        cnt = collections.Counter(nums)
        max_freq_num = max(cnt.values())
        return min(pos_map[n][1]-pos_map[n][0]+1 for n in cnt if cnt[n] == max_freq_num)

更精简的：

class Solution:
    def findShortestSubArray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        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)

注意：

Python 字典(Dictionary) setdefault()方法

---

描述

Python 字典 setdefault() 函数和get() 方法类似, 如果键不存在于字典中，将会添加键并将值设为默认值。

语法

setdefault()方法语法：

dict.setdefault(key, default=None)

参数

• key -- 查找的键值。
• default -- 键不存在时，设置的默认键值。

返回值

如果字典中包含有给定键，则返回该键对应的值，否则返回为该键设置的值。

实例

以下实例展示了 setdefault() 函数的使用方法：

实例(Python 2.0+)

#!/usr/bin/python # -*- coding: UTF-8 -*- dict = {'runoob': '菜鸟教程', 'google': 'Google 搜索'} print "Value : %s" % dict.setdefault('runoob', None) print "Value : %s" % dict.setdefault('Taobao', '淘宝')

以上实例输出结果为：

Value : 菜鸟教程
Value : 淘宝