295.find-median-from-data-stream

Statement

简体中文English

Metadata

Link: 数据流的中位数
Difficulty: Hard
Tag: 设计 双指针 数据流 排序 堆（优先队列）

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？

Metadata

Link: Find Median from Data Stream
Difficulty: Hard
Tag: Design Two Pointers Data Stream Sorting Heap (Priority Queue)

The median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value and the median is the mean of the two middle values.

For example, for arr = [2,3,4], the median is 3.
For example, for arr = [2,3], the median is (2 + 3) / 2 = 2.5.

Implement the MedianFinder class:

MedianFinder() initializes the MedianFinder object.
void addNum(int num) adds the integer num from the data stream to the data structure.
double findMedian() returns the median of all elements so far. Answers within 10^-5 of the actual answer will be accepted.

Example 1:

Input
["MedianFinder", "addNum", "addNum", "findMedian", "addNum", "findMedian"]
[[], [1], [2], [], [3], []]
Output
[null, null, null, 1.5, null, 2.0]
Explanation
MedianFinder medianFinder = new MedianFinder();
medianFinder.addNum(1);    // arr = [1]
medianFinder.addNum(2);    // arr = [1, 2]
medianFinder.findMedian(); // return 1.5 (i.e., (1 + 2) / 2)
medianFinder.addNum(3);    // arr[1, 2, 3]
medianFinder.findMedian(); // return 2.0

Constraints:

-10⁵ <= num <= 10⁵
There will be at least one element in the data structure before calling findMedian.
At most 5 * 10⁴ calls will be made to addNum and findMedian.

Follow up:

If all integer numbers from the stream are in the range [0, 100], how would you optimize your solution?
If 99% of all integer numbers from the stream are in the range [0, 100], how would you optimize your solution?

Solution

C++

#include <bits/stdc++.h>
using namespace std;
#define endl "\n"
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define mkp make_pair
#define all(x) (x).begin(), (x).end()
using db = double;
using ll = long long;
using ull = unsigned long long;
using pII = pair<int, int>;
using pLL = pair<ll, ll>;
constexpr int mod = 1e9 + 7;
template <class T1, class T2>
inline void chadd(T1 &x, T2 y, int Mod = mod) {
    x += y;
    while (x >= Mod) x -= Mod;
    while (x < 0) x += Mod;
}
template <class T1, class T2>
inline void chmax(T1 &x, T2 y) {
    if (x < y)
        x = y;
}
template <class T1, class T2>
inline void chmin(T1 &x, T2 y) {
    if (x > y)
        x = y;
}
inline int nextInt() {
    int x;
    cin >> x;
    return x;
}
void rd() {}
template <class T, class... Ts>
void rd(T &arg, Ts &...args) {
    cin >> arg;
    rd(args...);
}
#define dbg(x...)                             \
    do {                                      \
        cout << "\033[32;1m" << #x << " -> "; \
        err(x);                               \
    } while (0)
void err() {
    cout << "\033[39;0m" << endl;
}
template <class T, class... Ts>
void err(const T &arg, const Ts &...args) {
    cout << arg << ' ';
    err(args...);
}
template <template <typename...> class T, typename t, typename... A>
void err(const T<t> &arg, const A &...args) {
    for (auto &v : arg) cout << v << ' ';
    err(args...);
}
void ptt() {
    cout << endl;
}
template <class T, class... Ts>
void ptt(const T &arg, const Ts &...args) {
    cout << ' ' << arg;
    ptt(args...);
}
template <class T, class... Ts>
void pt(const T &arg, const Ts &...args) {
    cout << arg;
    ptt(args...);
}
void pt() {}
template <template <typename...> class T, typename t, typename... A>
void pt(const T<t> &arg, const A &...args) {
    for (int i = 0, sze = arg.size(); i < sze; ++i) cout << arg[i] << " \n"[i == sze - 1];
    pt(args...);
}
inline ll qpow(ll base, ll n) {
    assert(n >= 0);
    ll res = 1;
    while (n) {
        if (n & 1)
            res = res * base % mod;
        base = base * base % mod;
        n >>= 1;
    }
    return res;
}
// head
constexpr int N = 1e5 + 10;
int n;

class MedianFinder {
public:
    /** initialize your data structure here. */
    MedianFinder() {}

    priority_queue<int> big;
    priority_queue<int, vector<int>, greater<int>> low;

    void addNum(int num) {
        if (big.empty() && low.empty()) {
            big.push(num);
            return;
        }

        if (!big.empty()) {
            if (num <= big.top())
                big.push(num);
            else
                low.push(num);
        } else {
            if (num >= low.top())
                low.push(num);
            else
                big.push(num);
        }
    }

    double findMedian() {
        while (SZ(big) > SZ(low)) {
            low.push(big.top());
            big.pop();
        }

        while (SZ(low) - 1 > SZ(big)) {
            big.push(low.top());
            low.pop();
        }

        db res = 0;

        if (SZ(big) != SZ(low)) {
            res = low.top();
        } else {
            res = (big.top() + low.top()) * 1.0 / 2;
        }

        return res;
    }
};

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

#ifdef LOCAL

int main() {
    return 0;
}

#endif

最后更新: October 11, 2023