LeetCode Degree of an Array

原题链接在这里：https://leetcode.com/problems/degree-of-an-array/description/

题目：

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.

题解：

找出最大的frequency, 并标注对应的element出现过的左右index位置.

Time Complexity: O(n). n = nums.length.

Space: O(n).

AC Java:

 class Solution {
     public int findShortestSubArray(int[] nums) {
         HashMap<Integer, Integer> count = new HashMap<Integer, Integer>();
         HashMap<Integer, Integer> leftInd = new HashMap<Integer, Integer>();
         HashMap<Integer, Integer> rightInd = new HashMap<Integer, Integer>();
         int degree = 0;
 
         for(int i = 0; i<nums.length; i++){
             count.put(nums[i], count.getOrDefault(nums[i], 0)+1);
             degree = Math.max(degree, count.get(nums[i]));
 
             if(leftInd.get(nums[i]) == null){
                 leftInd.put(nums[i], i);
             }
             rightInd.put(nums[i], i);
         }
 
         int res = Integer.MAX_VALUE;
         for(int i = 0; i<nums.length; i++){
             if(count.get(nums[i]) == degree){
                 res = Math.min(res, rightInd.get(nums[i]) - leftInd.get(nums[i]) + 1);
             }
         }
 
         return res;
     }
 }