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.

解题思路：

把数组出现最多的保存到一个数字，对每个数字找头尾最先出现的序列，与min_distance做比较。

class Solution {
public:
    int findShortestSubArray(vector<int>& nums) {
        if(nums.size()==0) return 0;
        unordered_map<int,int> store;
        int max_digit=0;
        vector<int> digit;
        for(int i=0;i<nums.size();i++){
            store[nums[i]]++;

            if(store[nums[i]]>max_digit) max_digit = store[nums[i]];
        }
        if(max_digit == 1) return 1;
        for(unordered_map<int,int>::iterator iter= store.begin();iter!=store.end();iter++){
            if((*iter).second == max_digit) digit.push_back((*iter).first);
        }
        int min_distance = 99999;
        for(int i=0;i<digit.size();i++){
            //找最小间距
            int index1=0;
            int index2=0;
            for(int j=0;j<nums.size();j++){
               if(nums[j] == digit[i]) {index1 =j;break;}
            }
            for(int j=nums.size()-1;j>=0;j--){
               if(nums[j] == digit[i]) {index2 =j;break;}
            }
            if(index2-index1<min_distance) min_distance=index2-index1;

        }
        return min_distance+1;

    }
};