一、题目描述

中位数是有序列表中间的数。如果列表长度是偶数,中位数则是中间两个数的平均值。

例如,

[2,3,4] 的中位数是 3

[2,3] 的中位数是 (2 + 3) / 2 = 2.5

设计一个支持以下两种操作的数据结构:

void addNum(int num) - 从数据流中添加一个整数到数据结构中。
double findMedian() - 返回目前所有元素的中位数。
示例:

addNum(1)
addNum(2)
findMedian() -> 1.5
addNum(3)
findMedian() -> 2
进阶:

如果数据流中所有整数都在 0 到 100 范围内,你将如何优化你的算法?
如果数据流中 99% 的整数都在 0 到 100 范围内,你将如何优化你的算法?

二、题解

方法一:直接排序(会超时)

class MedianFinder {
public:
    vector<int> vec;
    /** initialize your data structure here. */
    MedianFinder() {

    }
    
    void addNum(int num) {
        vec.emplace_back(num);
    }
    
    double findMedian() {
        sort(vec.begin(),vec.end());
        int n = vec.size();
        return n&1?vec[n/2]:0.5*(vec[n/2-1]+vec[n/2]);
    }
};

/**
 * Your MedianFinder object will be instantiated and called as such:
 * MedianFinder* obj = new MedianFinder();
 * obj->addNum(num);
 * double param_2 = obj->findMedian();
 */

方法二:插入排序

class MedianFinder {
public:
    vector<int> vec;
    /** initialize your data structure here. */
    MedianFinder() {

    }
    
    void addNum(int num) {
        if(vec.empty()){
            vec.emplace_back(num);
        }else{
            vec.insert(lower_bound(vec.begin(),vec.end(),num),num);
        }

    }
    
    double findMedian() {
        int n = vec.size();
        return n&1?vec[n/2]:0.5*(vec[n/2-1]+vec[n/2]);
    }
};

方法三:两个堆实现

class MedianFinder {
public:
    priority_queue<int,vector<int>,less<int> > lo;
    priority_queue<int,vector<int>,greater<int> > hi;
public:
    /** initialize your data structure here. */
    MedianFinder() {

    }
    
    void addNum(int num) {
        lo.push(num);
        hi.push(lo.top());
        lo.pop();
        if(lo.size()<hi.size()){
            lo.push(hi.top());
            hi.pop();
        }
    }
    
    double findMedian() {
        return lo.size()>hi.size()?lo.top():((double)lo.top()+(double)hi.top())/2.0;
    }
};

/**
 * Your MedianFinder object will be instantiated and called as such:
 * MedianFinder* obj = new MedianFinder();
 * obj->addNum(num);
 * double param_2 = obj->findMedian();
 */

 方法四:使用Multiset和双指针

class MedianFinder {
public:
    multiset<int> data;
    multiset<int>::iterator lo_median, hi_median;
    /** initialize your data structure here. */
    MedianFinder():lo_median(data.end()),hi_median(data.end()) {

    }
    
    void addNum(int num) {
        int n = data.size();
        data.insert(num);
        if(n==0){
            lo_median = data.begin();
            hi_median = data.begin();
        }else if(n&1){//原来set大小是奇数,两指针指向同一元素,插入元素后应指向不同元素
            if(num<*lo_median){
                lo_median--;
            }else{
                hi_median++;
            }
        }else{//原来set大小是偶数,两指针指向相邻元素,插入元素后应指向相同元素
            if(num>*lo_median&&num<*hi_median){
                lo_median++;
                hi_median--;
            }else if(num>=*hi_median){//注意处理相等情况
                lo_median++;
            }else{
                lo_median=--hi_median;
            }
        }

    }
    
    double findMedian() {
        return (*lo_median+*hi_median)*0.5;
    }
};

/**
 * Your MedianFinder object will be instantiated and called as such:
 * MedianFinder* obj = new MedianFinder();
 * obj->addNum(num);
 * double param_2 = obj->findMedian();
 */

 

posted on 2021-02-03 15:22  曹婷婷  阅读(65)  评论(0编辑  收藏  举报