PAT A1029

最新推荐文章于 2025-08-21 12:02:17 发布

xiao1guaishou

最新推荐文章于 2025-08-21 12:02:17 发布

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分类专栏： two pointers PAT

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该博客主要介绍了PAT A1029题目的解题思路，涉及如何找到两个递增整数序列合并后的中位数。采用双指针法，在不额外存储完整序列的情况下，通过遍历两个序列找到中位数。

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题目
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 (≤2×105) 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:

题目大意
给出两个递增序列S1和S2,长度分别为N和M，求将他们合并成一个新的递增序列后的总位数

分析：双指针法
只需要输出合并后的序列的中位数即可
1.首先递增输入序列S1
2.在线处理S2，（如将S2输入到数组中，则会在最后一个测试点内存超出）
3.所以只要从小到大数到(n + m + 1) / 2的位置就行了~ count计总个数，给第一个数组设置指针i，每次从第二个数组中读入temp，检查第一个数组中前几个数是不是比temp小，小就count+1并判断是否到数了中间数，到了就输出。如果数完比temp小的数还没到中间数，count+1，检查temp是不是中间数，是就输出。循环上述过程。如果第二个数组读取完了，还没数到中间数，说明中间数在剩下的第一个数组中，就在剩下的数组中数到中间数位置即可

#include <cstdio>
#include <iostream>
const int INF =0x7fffffff;
const int Max = 1000100;
int main()
{
    int n1, n2;
    scanf("%d", &n1);
    int *A = new int[n1+1], temp;
    for(int i = 0; i < n1; i++){
        scanf("%d", &A[i]);
    }
    scanf("%d", &n2);
    int mid = (n1 + n2 - 1)/ 2;
    A[n1] = INF;
    int count1 = 0, i = 0, j = 0;
    scanf("%d", &temp);
    while(count1 < mid){
        if(A[i] <= temp)
            i++;
        else{
            j++;
            if(j < n2)
                scanf("%d", &temp);
            else if(j == n2)
                temp = INF;
            else break;
        }
        count1++;
    }
    printf("%d", A[i]<temp ? A[i] : temp);
    return 0;
}