697. Degree of an Array

Posted 2020-10-30 一日一更

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。

1. class Solution {  
2. public:  
3.     int findShortestSubArray(vector<int>& nums) {  
4.         if(nums.size()==0) return 0;  
5.         unordered_map<int,int> store;  
6.         int max_digit=0;  
7.         vector<int> digit;  
8.         for(int i=0;i<nums.size();i++){  
9.             store[nums[i]]++;  
10.               
11.             if(store[nums[i]]>max_digit) max_digit = store[nums[i]];  
12.         }  
13.         if(max_digit == 1) return 1;  
14.         for(unordered_map<int,int>::iterator iter= store.begin();iter!=store.end();iter++){  
15.             if((*iter).second == max_digit) digit.push_back((*iter).first);  
16.         }  
17.         int min_distance = 99999;  
18.         for(int i=0;i<digit.size();i++){  
19.             //找最小间距  
20.             int index1=0;  
21.             int index2=0;  
22.             for(int j=0;j<nums.size();j++){        
23.                if(nums[j] == digit[i]) {index1 =j;break;}  
24.             }  
25.             for(int j=nums.size()-1;j>=0;j--){        
26.                if(nums[j] == digit[i]) {index2 =j;break;}  
27.             }  
28.             if(index2-index1<min_distance) min_distance=index2-index1;  
29.               
30.         }  
31.         return min_distance+1;  
32.           
33.     }  
34. };