697. Degree of an Array - LeetCode

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.

Accepted

58,524

Submissions

113,812

Solution:

First Attempt: All test cases passed, but time limit exceeded.

class Solution {
    public int findShortestSubArray(int[] nums) {
     
        int start = 0; 
        int end = 0; 
        
        int max = Integer.MIN_VALUE;
        int val = 0;
                
         int len = Integer.MAX_VALUE; 
        
        
        int prev= Integer.MIN_VALUE;
        int prev2= Integer.MIN_VALUE;
        
        for(int i =0; i<nums.length; i++){
            
            if(prev2!=nums[i]&&Count(nums, nums[i])>max){
                
                max = Count(nums, nums[i]);
                  val = nums[i]; 
            }
            
            prev2= nums[i];
        }
        
       // System.out.println("val = "+ val+ " max = "+ max);
        
        for(int i = 0; i<nums.length; i++){
            
            if(nums[i]!=prev && Count(nums, nums[i])==max){
            int tmp1_start = 0; 
            int tmp2_end = 0;
                for(int j = 0; j<nums.length; j++){
                    
                    if(nums[j]==nums[i]){
                        tmp1_start = j;   
                        break;
                    }
                    // System.out.println(nums[i]+" "+ "j = "+ j);
                  
                }
                
                for(int k = nums.length-1; k>=0; k--){
                    if(nums[k]==nums[i]){
                        tmp2_end = k;
                        break;
                    }
                     //System.out.println("  k  "+ k );
                    
                }
                
                if((tmp2_end - tmp1_start)+1 <len){
                    len = (tmp2_end - tmp1_start)+1; 
                }
            }
            prev = nums[i];
        }
        return len;
    }
    
    static int Count(int[] nums, int cur){
        int count = 0; 
        for(int i = 0; i < nums.length; i++){
            
            if(nums[i] == cur){
                count++;
            }
        }
        
        return count;

    }
    
}

原文地址：https://www.cnblogs.com/codingyangmao/p/11555980.html