leetcode-697. Degree of an Array

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.

思路：

肯定很容易想到 先找到数组里出现次数最多的数，在用最后一次出现的位置减去第一次出现的位置，还要考虑到又出现频率一样的数，怎么计算度，问题就是怎么组织数据来实现

一开始我也想到这方法，但是却不知道怎么组织数据，一开始用数组来存元素的次数，用数组来计算度，但是遇到次数一致的就无法解决。而且用数组来存次数也需要开一个很大的数组也考虑了用map

后来看了discuss，思路是一样的，但是用的map<Integer,int[]>，自己没想到

class Solution {
     public int findShortestSubArray(int[] nums) {
        if (nums.length == 0 || nums == null) return 0;
        Map<Integer, int[]> map = new HashMap<>();
        for (int i = 0; i < nums.length; i++){
           if (!map.containsKey(nums[i])){
               map.put(nums[i], new int[]{1, i, i});  // the first element in array is degree, second is first index of this key, third is last index of this key
           } else {
               int[] temp = map.get(nums[i]);
               temp[0]++;
               temp[2] = i;
           }
        }
        int degree = Integer.MIN_VALUE, res = Integer.MAX_VALUE;
        for (int[] value : map.values()){
            if (value[0] > degree){
                degree = value[0];
                res = value[2] - value[1] + 1;
            } else if (value[0] == degree){
                res = Math.min( value[2] - value[1] + 1, res);
            } 
        }
        return res;
    }
}

其中的

               int[] temp = map.get(nums[i]);
               temp[0]++;
               temp[2] = i;

是能带改变map里的value的值的。。