标签：frequency exp his bar max nbsp span amp contain
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.
1 class Solution { 2 public int findShortestSubArray(int[] nums) { 3 // maps number with [count, startIndex, endIndex] 4 HashMap<Integer, int[]> map = new HashMap<Integer, int[]>(); 5 6 int result = Integer.MAX_VALUE, maxCount = 0; 7 for (int i = 0; i < nums.length; i++) { 8 int[] numData = new int[3]; 9 if (map.containsKey(nums[i])) { 10 int[] temp = map.get(nums[i]); 11 numData[0] = temp[0] + 1; 12 numData[1] = temp[1]; 13 } else { 14 numData[0] = 1; 15 numData[1] = i; 16 } 17 numData[2] = i; 18 map.put(nums[i], numData); 19 20 if (numData[0] > maxCount) { 21 maxCount = numData[0]; 22 result = numData[2] - numData[1] + 1; 23 } else if (numData[0] == maxCount) { 24 result = Math.min(result, numData[2] - numData[1] + 1); 25 } 26 } 27 28 return result; 29 } 30 }
标签：frequency exp his bar max nbsp span amp contain
原文地址：https://www.cnblogs.com/amazingzoe/p/9082244.html
