697. Degree of an Array

最新推荐文章于 2025-05-04 07:15:00 发布

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最新推荐文章于 2025-05-04 07:15:00 发布

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本文介绍了一种算法，用于寻找具有与整个数组相同最大频率元素的最短连续子数组长度。通过使用三个哈希映射来跟踪每个元素的首次出现、末次出现及出现次数，进而确定满足条件的最短子数组。

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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.

这个题目是叫我们求数组中最小子数组的长度

	public static void main(String[] args) {
		// 此题中key用来当作数组中的数字 value当作数组下标 {1, 2, 3, 2, 5}
		HashMap<Integer, Integer> left = new HashMap<>();
		HashMap<Integer, Integer> right = new HashMap<>();
		HashMap<Integer, Integer> count = new HashMap<>();
		
		int x = 0;
		
		int num[] = {1, 4, 3, 4, 5};
		for (int i=0; i<num.length; i++) {
			x = num[i];
			//这里判断left中是否已经有这个元素了，如果没有存到left和right里去 有的话就存进right中
			//这样存主要目的是让left内存储第一次出现这个元素的下标，而right存储这个元素最后一次出现的下标
			if (left.get(x) == null)
				left.put(x, i);
			right.put(x, i);
			count.put(x, count.getOrDefault(x, 0) + 1); // count这个map用来存element在数组中的出现次数 
		}
		
		int degree = Collections.max(count.values()); // 找出数组中出现次数最多的数的次数
		int ans = num.length;
		
		// 增强型for循环 遍历count的key集合
		for (int i : count.keySet()) {
			if (count.get(i) == degree) {
				ans = Math.min(ans, right.get(i) - left.get(i) + 1);
			}
		}
		
		System.out.println(ans);
	}