[LeetCode] 697. Degree of an Array

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.

数组的度。

给定一个非空且只包含非负数的整数数组 nums，数组的度的定义是指数组里任一元素出现频数的最大值。

你的任务是在 nums 中找到与 nums 拥有相同大小的度的最短连续子数组，返回其长度。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/degree-of-an-array
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路是用两个hashmap，一个记录数字的出现次数，一个记录每个不同数字最早出现的坐标。当每次发现一个出现次数最多的数字的时候，就更新子数组的长度；如果发现有多个数字的出现次数都是最多的情况下，则查看是否能缩小子数组的长度。

时间O(n)

空间O(n)

Java实现

class Solution {
    public int findShortestSubArray(int[] nums) {
        Map<Integer, Integer> count = new HashMap<>();
        Map<Integer, Integer> first = new HashMap<>();
        int res = 0;
        int degree = 0;
        for (int i = 0; i < nums.length; i++) {
            first.putIfAbsent(nums[i], i);
            count.put(nums[i], count.getOrDefault(nums[i], 0) + 1);
            if (count.get(nums[i]) > degree) {
                degree = count.get(nums[i]);
                res = i - first.get(nums[i]) + 1;
            } else if (count.get(nums[i]) == degree) {
                res = Math.min(res, i - first.get(nums[i]) + 1);
            }
        }
        return res;
    }
}

LeetCode 题目总结