697. Degree of an Array
Easy
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: nums = [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: nums = [1,2,2,3,1,4,2] Output: 6 Explanation: The degree is 3 because the element 2 is repeated 3 times. So [2,2,3,1,4,2] is the shortest subarray, therefore returning 6.
Constraints:
nums.length
will be between 1 and 50,000.nums[i]
will be an integer between 0 and 49,999.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 | class Solution { public int findShortestSubArray(int[] nums) { int n = nums.length; int[][] map = new int[50000][3];//startIndex, endIndex, count int ret = n, degree = 0; for(int i=0; i<n; ++i){ int v = nums[i]; if(map[v][2]==0) map[v][0] = i;//startIndex map[v][1] = i;//endIndex map[v][2]++;//count if(map[v][2]>degree){ degree = map[v][2]; ret = map[v][1]-map[v][0]+1; }else if(map[v][2]==degree && map[v][1]-map[v][0]+1<ret){ ret = map[v][1]-map[v][0]+1; } } return ret; } } |
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