[编程入门]链表合并-题解(C++代码)

合并链表之后再排序
原本的思路是将两个有序的链表来回引用指针排序
但是不会
注释的语句是有序链表合并的源代码

#include<iostream>
using namespace std;
struct a {
    int number;
    int achievement;
};
struct b {
    int number;
    int achievement;
};
struct c {
    int number;
    int achievement;
};
int main()
{
    int n_int;
    int m_int;
    cin >> n_int;
    cin >> m_int;
    a* p = new a[n_int];
    for (int i_for1_int = 0; i_for1_int < n_int; i_for1_int++) {
        cin >> (*(p + i_for1_int)).number;
        cin >> (*(p + i_for1_int)).achievement;
    }
    b* q = new b[m_int];
    for (int i_for1_int = 0; i_for1_int < m_int; i_for1_int++) {
        cin >> (*(q + i_for1_int)).number;
        cin >> (*(q + i_for1_int)).achievement;
    }
    /*for (int i_for1_int = 0; i_for1_int < (n_int>m_int ? n_int :m_int); i_for1_int++) {
        if (i_for1_int < n_int) {
            for (int i_for2_int = 0; i_for2_int < n_int; i_for2_int++) {
                if ((*(p + i_for1_int)).number > (*(p + i_for2_int)).number) {
                    int change_int=(*(p + i_for1_int)).number;
                    (*(p + i_for1_int)).number = (*(p + i_for2_int)).number;
                    (*(p + i_for2_int)).number = change_int;
                    change_int = (*(p + i_for1_int)).achievement;
                    (*(p + i_for1_int)).achievement = (*(p + i_for2_int)).achievement;
                    (*(p + i_for2_int)).achievement = change_int;
                }
            }
        }
        if (i_for1_int < m_int) {
            for (int i_for2_int = 0; i_for2_int < m_int; i_for2_int++) {
                if ((*(q + i_for1_int)).number > (*(q + i_for2_int)).number) {
                        int change_int = (*(q + i_for1_int)).number;
                        (*(q + i_for1_int)).number = (*(q + i_for2_int)).number;
                        (*(q + i_for2_int)).number = change_int;
                        change_int = (*(q + i_for1_int)).achievement;
                        (*(q + i_for1_int)).achievement = (*(q + i_for2_int)).achievement;
                        (*(q + i_for2_int)).achievement = change_int;
                }
            }
        }
    }*/
    c* v = new c[m_int + n_int];
    for (int i_for1_int = 0; i_for1_int < n_int+m_int; i_for1_int++) {
        if (i_for1_int < n_int) {
            (*(v + i_for1_int)).number = (*(p + i_for1_int)).number;
            (*(v + i_for1_int)).achievement = (*(p + i_for1_int)).achievement;
        }else{
            (*(v + i_for1_int)).number = (*(q + i_for1_int-n_int)).number;
            (*(v + i_for1_int)).achievement = (*(q + i_for1_int-n_int)).achievement;
        }
    }
    for (int i_for1_int = 0; i_for1_int < n_int + m_int; i_for1_int++) {
        for (int i_for2_int = 0; i_for2_int < n_int + m_int; i_for2_int++) {
            if ((*(v + i_for1_int)).number < (*(v + i_for2_int)).number) {
                int change_int = (*(v + i_for1_int)).number;
                (*(v + i_for1_int)).number = (*(v + i_for2_int)).number;
                (*(v + i_for2_int)).number = change_int;
                change_int = (*(v + i_for1_int)).achievement;
                (*(v + i_for1_int)).achievement = (*(v + i_for2_int)).achievement;
                (*(v + i_for2_int)).achievement = change_int;
            }
        }
    }
    /*for (int i_for1_int = 0; i_for1_int < m_int+n_int; i_for1_int++) {
        for (int i_for2_int = 0; i_for2_int < n_int; i_for2_int++) {
            for (int i_for3_int = 0; i_for3_int < m_int; i_for3_int++) {
                if ((*(p + i_for2_int)).number < (*(q + i_for3_int)).number) {
                    (*(v + i_for1_int)).number = (*(p + i_for2_int)).number;
                    (*(v + i_for1_int)).achievement = (*(p + i_for2_int)).achievement;
                }
                else {
                    (*(v + i_for1_int)).number = (*(q + i_for3_int)).number;
                    (*(v + i_for1_int)).achievement = (*(q + i_for3_int)).achievement;
                }
            }
        }
    }*/
    for (int i_for1_int = 0; i_for1_int < n_int+m_int; i_for1_int++) {
        cout << (*(v + i_for1_int)).number << " " << (*(v + i_for1_int)).achievement << endl;
    }
}