［LeetCode］Degree of an Array 数组的

链接：https://leetcode.com/problems/degree-of-an-array/description/
难度：Easy
题目：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.

翻译：给定一个非空非负的整型数组，定义数组的度为数组中元素出现的最大次数。任务是找出度和数组的度相同的最小子串

思路：记录下第一次出现和最后一次出现的位置就好了，两者相减就是最短长度。对于有多个出现次数最多元素的情况，只需要找出这些元素的最短子串中最小的就好了。

参考代码：
Java

class Solution {
    public int findShortestSubArray(int[] nums) {
        Map<Integer, Integer> left = new HashMap();
        Map<Integer, Integer> right = new HashMap();
        Map<Integer, Integer> count = new HashMap();
        
        for(int i=0; i < nums.length; i++ ){
            if(!left.containsKey(nums[i]))
                left.put(nums[i], i);
            right.put(nums[i], i);
            count.put(nums[i], count.getOrDefault(nums[i],0)+1);
        }
        
        int degree = Collections.max(count.values());
        int length = Integer.MAX_VALUE;
        for(int i=0; i<nums.length; i++){
            if(count.get(nums[i])==degree){
                length = Math.min(length, right.get(nums[i]) - left.get(nums[i]) + 1);
            }
        }
        return length;
    }   
}