# 697.Degree of an Array

**697.Degree of an Array**

难度:Easy

> 给定一个非空且只包含非负数的整数数组 nums, 数组的度的定义是指数组里任一元素出现频数的最大值。

你的任务是找到与 nums 拥有相同大小的度的最短连续子数组，返回其长度。

```
示例 1:

输入: [1, 2, 2, 3, 1]
输出: 2
解释: 
输入数组的度是2，因为元素1和2的出现频数最大，均为2.
连续子数组里面拥有相同度的有如下所示:
[1, 2, 2, 3, 1], [1, 2, 2, 3], [2, 2, 3, 1], [1, 2, 2], [2, 2, 3], [2, 2]
最短连续子数组[2, 2]的长度为2，所以返回2.
示例 2:

输入: [1,2,2,3,1,4,2]
输出: 6
注意:

nums.length 在1到50,000区间范围内。
nums[i] 是一个在0到49,999范围内的整数。
```

创建一个Hash表，然后找出频次最多的元素，找到元素的起始索引，然后找出所有频次最多元素的保留度的最短长度，求出其最小值。

```
class Solution {
int len(vector<int >nums, int index, int degree)
{
    int start=index+1;
    int count=1;
    while(count<degree)
    {
        if(nums[start]==nums[index])
        {
            ++count;

        }
        ++start;
    }
    return start-index ;
}
public:
    int findShortestSubArray(vector<int>& nums) {
        unordered_map<int,int>degree_of_nums;
        int max_index=nums[0];
        for(auto n:nums)
        {
            degree_of_nums[n]++;
            if(degree_of_nums[max_index]<degree_of_nums[n])
                max_index=n;

    }
        int degree=degree_of_nums[max_index];
        max_index=find(nums.begin(),nums.end(), max_index) - nums.begin();

        //cout<<max_index<< "--" << degree_of_nums[nums[max_index]] <<endl;
    int res=len(nums, max_index, degree_of_nums[nums[max_index]]);
        for(auto &e: degree_of_nums)
        {
            if(e.first != nums[max_index] && e.second == degree)
                res = min(res, len(nums, find(nums.begin(), nums.end(), e.first)-nums.begin(), degree));
        }
        return res;
    }
};
```

> 执行用时 :348 ms, 在所有 C++ 提交中击败了5.49%的用户\
> 内存消耗 :284MB, 在所有 C++ 提交中击败了5.62%的用户


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