# 697. Degree of an Array

&#x20;[**697. Degree of an Array**](https://leetcode.com/problems/degree-of-an-array/description/)

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.

***My Solutions：***

用一个Hashmap存上key为数字，value为int【】的数组，数组储存了最前的index，最后的index，和key的个数。

```java
class Solution {
    public int findShortestSubArray(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        
        Map<Integer, int[]> map = new HashMap<>();
        int degree = 1;
        
        for (int i = 0; i < nums.length; i++) {
            if (!map.containsKey(nums[i])) {
                map.put(nums[i], new int[]{i, i, 1}); //start index, end index, frequency
            } else {
                int[] vars = map.get(nums[i]);
                vars[1] = i; //更新end index
                vars[2]++; //更新frequency
                degree = Math.max(degree, vars[2]); //更新degree
            }
        }
        
        int res = nums.length;
        for (int[] values : map.values()) {
            if (values[2] == degree) {
                res = Math.min(res, values[1] - values[0] + 1); //values[1] - values[0] + 1 是subarray的长度
            }
        }
        return res;
    }
}
```
