697. Degree of an Array

#### QUESTION:

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.

#### EXPLANATION:

1.遍历数组

2.找到相同的数字的个数以及最大的距离。

3.遍历结果，找到最大个数和最小距离。

#### SOLUTION:

``````class Solution {
class ShortestSubArray{
int firstIndex = -1;
int count = 1;
int length = 1;
}
public int findShortestSubArray(int[] nums) {
HashMap<Integer,ShortestSubArray> map = new HashMap<>();
for(int i = 0;i<nums.length;i++){
ShortestSubArray tmp = map.get(nums[i]);
if(tmp ==null){
tmp = new ShortestSubArray();
tmp.firstIndex = i;
}else{
tmp.length = i-tmp.firstIndex+1;
tmp.count++;
}
map.put(nums[i],tmp);
}
int result = Integer.MAX_VALUE;
int count = 0;
Iterator<Integer> iterator = map.keySet().iterator();
while (iterator.hasNext()){
Integer next = iterator.next();
ShortestSubArray shortestSubArray = map.get(next);
if(shortestSubArray.count>count){
count = shortestSubArray.count;
result = shortestSubArray.length;
}else if(shortestSubArray.count==count&&shortestSubArray.length<result){
count = shortestSubArray.count;
result = shortestSubArray.length;
}
}
return result;
}
}
``````
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