标签：you hash element style get logs int out hashmap
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:
1 class Solution { 2 public int findShortestSubArray(int[] nums) { 3 HashMap<Integer, Integer> count = new HashMap<Integer, Integer>(); 4 HashMap<Integer, Integer> leftInd = new HashMap<Integer, Integer>(); 5 HashMap<Integer, Integer> rightInd = new HashMap<Integer, Integer>(); 6 int degree = 0; 7 8 for(int i = 0; i<nums.length; i++){ 9 count.put(nums[i], count.getOrDefault(nums[i], 0)+1); 10 degree = Math.max(degree, count.get(nums[i])); 11 12 if(leftInd.get(nums[i]) == null){ 13 leftInd.put(nums[i], i); 14 } 15 rightInd.put(nums[i], i); 16 } 17 18 int res = Integer.MAX_VALUE; 19 for(int i = 0; i<nums.length; i++){ 20 if(count.get(nums[i]) == degree){ 21 res = Math.min(res, rightInd.get(nums[i]) - leftInd.get(nums[i]) + 1); 22 } 23 } 24 25 return res; 26 } 27 }
标签：you hash element style get logs int out hashmap
原文地址：http://www.cnblogs.com/Dylan-Java-NYC/p/7749867.html
