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

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.
 

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.

Solution

此題的意思是在陣列中找到數字的最高頻率作為degree，然後在此陣列中找到一個最短的子陣列使它擁有相同的degree，返回該子陣列的長度。也就是說，在子陣列中必須包含出現頻率最高的所有數字中的至少一個數字，並且這個數字全部都在子陣列中。因此，我們就需要記錄數字的第一次出現位置與最後一次出現位置，位置差最短的子陣列即為目標子陣列。

以下程式碼使用了三個map，分別記錄數字第一次出現的位置、數字最後一次出現位置與數字的出現次數。

class Solution {
    public int findShortestSubArray(int[] nums) {
        HashMap<Integer, Integer> left = new HashMap<>();
        HashMap<Integer, Integer> right = new HashMap<>();
        HashMap<Integer, Integer> count =
 
 new HashMap<>();
        for(int i = 0; i < nums.length; i++) {
        	int num = nums[i];
        	if(left.get(num) == null)
        		left.put(num, i);
        	right.put(num, i);
        	count.put(num, count.getOrDefault(num, 0)+1);
        }
        
        int degree = Collections.max(count.values());
        int shortest = Integer.MAX_VALUE;
        for(int num : count.keySet()) {
        	if(count.get(num) == degree) {
        		shortest = Math.min(shortest, right.get(num) - left.get(num) + 1);
        	}
        }
        return shortest;
    }
}