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.class Solution { public int findShortestSubArray(int[] nums) { Map<Integer, Integer> hmap = new HashMap<>(); int degree = -1; //int number = nums[0]; for(int num : nums){ int count = hmap.getOrDefault(num, 0) + 1; if(degree < count){ degree = count; //number = num; } hmap.put(num, count); } int minVal = Integer.MAX_VALUE; for(Map.Entry<Integer, Integer> item : hmap.entrySet()){ if(item.getValue() == degree){ int number = item.getKey(); int left = -1; int right = -1; int len = nums.length; for(int i=0; i<len; i++){ if(nums[i] == number && left == -1){ left = i; } if(nums[len-1-i] == number && right == -1){ right = len - 1 - i; } } len = right < left ? 0 : right - left + 1; if(minVal > len){ minVal = len; } } } return minVal; } }
[문제] https://leetcode.com/problems/degree-of-an-array/