697. Degree of an Array

最新推荐文章于 2025-05-04 07:15:00 发布

BigFatSheep

最新推荐文章于 2025-05-04 07:15:00 发布

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分类专栏： Leetcode Easy

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本文介绍了一种算法，用于找出给定数组中具有相同度（即最大元素频率）的最短连续子序列。通过两次遍历并使用哈希映射，文章详细解释了如何确定子序列的长度，并提供了具体示例。

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Description

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.

Problem URL

Solution

一个数组中出现最多的元素的频率称为数组的度，找到数组中包括了这个度的连续子序列的最短长度。

We use a hash map to contain the frequence of each element in the array, then get the max frequence by a for loop.

After get the max frequence, we iterate map.keySet() for another time, find the element who have the max frequence. Then using two pointers to search from head and tail, and get the subarray, count it length.

Code

class Solution {
    public int findShortestSubArray(int[] nums) {
        if(nums == null || nums.length == 0){
            return 0;
        }
        Map<Integer, Integer> map = new HashMap<>();
        for (int n : nums){
            map.put(n, map.getOrDefault(n, 0) + 1);
        }
        int max = Integer.MIN_VALUE;
        for (Integer n : map.keySet()){
            max = Math.max(max, map.get(n));
        }
        int res = Integer.MAX_VALUE;
        for (Integer n : map.keySet()){
            if (map.get(n) == max){
                int i = 0, j = nums.length - 1;
                while (i < j){
                    if (nums[i] == n && nums[j] == n){
                        break;
                    }
                    if (nums[i] != n){
                        i++;
                    }
                    if (nums[j] != n){
                        j--;
                    }
                }
                res = Math.min(res, j - i + 1);
            }
        }
        return res;
    }
}