[剑指offer]面试题第[41]题[Leetcode][第235题][JAVA][数据流中的中位数][优先队列][堆]

【问题描述】[困难]

【解答思路】

1. 思路1

时间复杂度：O(logN) 空间复杂度：O(N)

import java.util.PriorityQueue;public class MedianFinder {/*** 当前大顶堆和小顶堆的元素个数之和*/private int count;private PriorityQueue<Integer> maxheap;private PriorityQueue<Integer> minheap;/*** initialize your data structure here.*/public MedianFinder() {count = 0;//maxHeap = new PriorityQueue<>(Collections.reverseOrder());maxheap = new PriorityQueue<>((x, y) -> y - x);minheap = new PriorityQueue<>();}public void addNum(int num) {count += 1;maxheap.offer(num);minheap.add(maxheap.poll());// 如果两个堆合起来的元素个数是奇数，小顶堆要拿出堆顶元素给大顶堆if ((count & 1) != 0) {maxheap.add(minheap.poll());}/*//如果不平衡则调整 不用countif (minHeap.size() > maxHeap.size()) {maxHeap.offer(minHeap.poll());}*/}public double findMedian() {if ((count & 1) == 0) {// 如果两个堆合起来的元素个数是偶数，数据流的中位数就是各自堆顶元素的平均值return (double) (maxheap.peek() + minheap.peek()) / 2;} else {// 如果两个堆合起来的元素个数是奇数，数据流的中位数大顶堆的堆顶元素return (double) maxheap.peek();}
//不用count
public double findMedian() {if (maxHeap.size() == minHeap.size()) {return (maxHeap.peek() + minHeap.peek()) * 0.5;}return maxHeap.peek();}}
}

2. 思路2

复杂度

class MedianFinder {Queue<Integer> A, B;public MedianFinder() {A = new PriorityQueue<>(); // 小顶堆，保存较大的一半B = new PriorityQueue<>((x, y) -> (y - x)); // 大顶堆，保存较小的一半}public void addNum(int num) {if(A.size() != B.size()) {A.add(num);B.add(A.poll());} else {B.add(num);A.add(B.poll());}}public double findMedian() {return A.size() != B.size() ? A.peek() : (A.peek() + B.peek()) / 2.0;}
}

【总结】

1.最大堆/最小堆初始化

最小堆
PriorityQueue minheap = new PriorityQueue<>();
最大堆
PriorityQueue maxheap = new PriorityQueue<>((x, y) -> y - x);
PriorityQueue maxHeap = new PriorityQueue<>(Collections.reverseOrder());

2.找规律模拟过程不可能是一蹴而就而是多次失败后才能找到的规律和转移过程

转载链接：https://leetcode-cn.com/problems/find-median-from-data-stream/solution/you-xian-dui-lie-python-dai-ma-java-dai-ma-by-liwe/

参考链接：https://leetcode-cn.com/problems/shu-ju-liu-zhong-de-zhong-wei-shu-lcof/solution/you-xian-dui-lie-wu-fei-hua-jian-dan-yi-dong-by-je/

参考链接：https://leetcode-cn.com/problems/shu-ju-liu-zhong-de-zhong-wei-shu-lcof/solution/mian-shi-ti-41-shu-ju-liu-zhong-de-zhong-wei-shu-y/