PAT (Advanced Level) Practice 1029 Median（25分）【双指针】

Given an increasing sequence S of N integers, the median is the number at the middle position. For example, the median of S1 = { 11, 12, 13, 14 } is 12, and the median of S2 = { 9, 10, 15, 16, 17 } is 15. The median of two sequences is defined to be the median of the nondecreasing sequence which contains all the elements of both sequences. For example, the median of S1 and S2 is 13.

Given two increasing sequences of integers, you are asked to find their median.

Input Specification:

Each input file contains one test case. Each case occupies 2 lines, each gives the information of a sequence. For each sequence, the first positive integer $N(\le 2\times 10^5)$ is the size of that sequence. Then N integers follow, separated by a space. It is guaranteed that all the integers are in the range of long int.

Output Specification:

For each test case you should output the median of the two given sequences in a line.

Sample Input:

4 11 12 13 14
5 9 10 15 16 17

Sample Output:

13

题意

给定两个有序数组，计算二者合并后的中位数。

思路

两个数组都读下来查找会内存超限，需要一边读第二个数组一边找中位数。而且有可能第二个数组已经读完了但是还没有找到中位数的位置。

代码

#include 

using namespace std;

long int arr[200005];

int main() {
    cin.tie(nullptr);
    cout.tie(nullptr);
    ios::sync_with_stdio(false);

    int n1, n2, cur1 = 0, cur2 = 0, mid, ans, t;

    cin >> n1;  // 读入第一个数组
    for (int i = 0; i < n1; ++i)
        cin >> arr[i];

    cin >> n2;
    mid = (n1 + n2 + 1) / 2;

    while (cur1 + cur2 < mid and cur2 < n2) {  // 在线处理第二个数组
        cin >> t;  // 注意最多读入n2个数
        while (cur1 < n1 and cur1 + cur2 < mid and arr[cur1] <= t)
            ans = arr[cur1++];
        if (cur1 + ++cur2 <= mid) ans = t;
    }

    while (cur1 + cur2 < mid)  // n2个数全部读入但还是没有到达mid
        ans = arr[cur1++];

    cout << ans;
}