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: 2Explanation: 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：
maxs记录元素的最大下标，mins记录元素的最小下标，cnts记录元素的出现个数，三者均是HashMap结构；
通过遍历记录元素与上三个对应的记录，只要找到最大次数对应的元素，则就找到了该元素的最大和最小下标，即找到了度相同的最小子数组。
时间复杂度：O(n)；空间复杂度：O(1)

class Solution {    public int findShortestSubArray(int[] nums) {        Map<Integer, Integer> start = new HashMap<Integer, Integer>();        Map<Integer, Integer> end = new HashMap<Integer, Integer>();        Map<Integer, Integer> count  = new HashMap<Integer, Integer>();        int maxCount = 0;        for (int i = 0; i < nums.length; i++){            if (!count.containsKey(nums[i])){                count.put(nums[i], 0);                start.put(nums[i], i);            }            count.put(nums[i], count.get(nums[i])+1);            end.put(nums[i], i);            maxCount = Math.max(maxCount, count.get(nums[i]));        }        int minLength = Integer.MAX_VALUE;        for (Integer key : count.keySet()){            if (count.get(key) == maxCount){                minLength = Math.min(minLength,end.get(key)-start.get(key)+1);            }        }        return minLength;    }}