leetcode 697
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
给定非空整数nums的非空数组,该数组的度数被定义为其任何一个元素的最大频率。
你的任务是找到num的(连续的)子阵列的最小可能长度,其与nums具有相同的度数。
例1:
输入:[1,2,2,3,1]
输出:2
说明:
输入数组的度数为2,因为元素1和2都出现两次。
在具有相同程度的子阵列中:
[1,2,2,3,1],[1,2,2,3],[2,2,3,1],[1,2,2],[2,2,3],[2] ,2]
最短的长度是2.所以返回2。
例2:
输入:[1,2,2,3,1,4,2]
输出:6
class Solution { public int findShortestSubArray(int[] nums) { int maxcount = 1; HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>(); for (int i : nums) { if (hm.containsKey(i)) { hm.put(i, hm.get(i) + 1); if (maxcount < hm.get(i)) { maxcount = hm.get(i); } } else { hm.put(i, 1); } } Set<Integer> set = hm.keySet(); int minlength = Integer.MAX_VALUE; for (int s : set) { int temp = Integer.MAX_VALUE; if (hm.get(s) == maxcount) { int i = 0, j = nums.length - 1; while (nums[i] != s && i < j) i++; while (nums[j] != s && i < j) j--; temp = j - i + 1; } minlength = Math.min(temp, minlength); } return minlength; } }