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最新推荐文章于 2024-08-05 21:31:54 发布

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分类专栏： leetcode 文章标签： LeetCode heap

本文链接：https://blog.youkuaiyun.com/xw_Summer/article/details/86619104

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本文详细介绍了LeetCode上两道经典算法题的解决方案，包括寻找数组中第K大的元素及找出数组中最频繁出现的K个元素。通过桶排序等高效算法，实现优于O(nlogn)的时间复杂度。

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215. Kth Largest Element in an Array

Find the kth largest element in an unsorted array. Note that it is the
kth largest element in the sorted order, not the kth distinct element.

Note:
You may assume k is always valid, 1 ≤ k ≤ array’s length.

https://leetcode.com/problems/kth-largest-element-in-an-array/discuss/60294/Solution-explained

347. Top K Frequent Elements

Given a non-empty array of integers, return the k most frequent elements.

Note:

You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
Your algorithm’s time complexity must be better than O(n log n), where n is the array’s size.

Java O(n) Solution - Bucket Sort:

Idea is simple. Build a array of list to be buckets with length 1 to sort.

class Solution {
    public List<Integer> topKFrequent(int[] nums, int k) {

        List<Integer>[] bucket = new List[nums.length + 1];
        Map<Integer, Integer> frequencyMap = new HashMap<Integer, Integer>();

        for (int n : nums) {
            frequencyMap.put(n, frequencyMap.getOrDefault(n, 0) + 1);
        }

        for (int key : frequencyMap.keySet()) {
            int frequency = frequencyMap.get(key);
            if (bucket[frequency] == null) {
                bucket[frequency] = new ArrayList<>();
            }
            bucket[frequency].add(key);
        }

        List<Integer> res = new ArrayList<>();

        for (int pos = bucket.length - 1; pos >= 0 && res.size() < k; pos--) {
            if (bucket[pos] != null) {
                res.addAll(bucket[pos]);
            }
        }
        return res;
    }
}