7-52 两个有序链表序列的交集 (20分)

≡放荡痞女 提交于 2019-12-17 22:58:46

7-52 两个有序链表序列的交集 (20分)

AC代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;
typedef struct node{
    int data;
    struct node *nxt;
}node, *linklist;
linklist Read(){
    linklist L, op;
    L = (linklist)malloc(sizeof(node));
    L->data = -1;
    L->nxt = NULL;
    op = L;
    int num;
    while (scanf("%d", &num) && num != -1){
        linklist p = (linklist)malloc(sizeof(node));
        p->data = num;
        p->nxt = NULL;
        op->nxt = p;
        op = op->nxt;
    }
    op->nxt = NULL;
    return L;
}
linklist Common(linklist la, linklist lb){
    linklist L, op, pa, pb;
    L = (linklist)malloc(sizeof(node));
    L->data = -1;
    L->nxt = NULL;
    op = L;
    pa = la->nxt;
    pb = lb->nxt;
    while (pa != NULL && pb != NULL){
        if (pa->data == pb->data){
            op->nxt = pa;
            pa = pa->nxt;
            pb = pb->nxt;
            op = op->nxt;
        }
        else if (pa->data < pb->data) pa = pa->nxt;
        else pb = pb->nxt;
    }
    op->nxt = NULL;
    return L;
}
int main(){
    linklist la, lb, lc;
    la = Read();
    lb = Read();
    lc = Common(la, lb);
    lc = lc->nxt;
    int flag = 0;
    if (lc == NULL) printf("NULL");
    else {
        while (lc != NULL){
            if (flag != 0) printf(" ");
            printf("%d", lc->data);
            flag = 1;
            lc = lc->nxt;
        }
    }
    return 0;
}
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