递增链表的交集

题目：

已知两个递增链表LA,LB分别表示集合A,B，设计算法实现，求集合A,B的交集。用递增的方式存储链表LC，并且求解时间尽可能地小。

时间复杂度为两个链表数目之和O（m+n）：

// IntersectionLink.cpp : 定义控制台应用程序的入口点。 //算法描述，la，lb递增链表，lc为两者的交集，也是递增的。 #include <stdio.h> #include<stdlib.h> typedef struct node { int data; struct node* next; }Link; //链表初始化 必须建立递增的链表 Link* CreateLa(Link *head,int n) { Link *p; //头插法 //for(int i=0;i<n;i++) //{ // p = (Link *)malloc(sizeof(Link)); // p->data = 2*i; // p->next = head->next; // head->next = p; //} //尾插法 for(int i=0;i<n;i++) { p = (Link *)malloc(sizeof(Link)); p->data = 2*i; head->next = p; head = p; } head->next = NULL; return head; } Link* CreateLb(Link *head,int n) { Link *p; //尾插法 for(int i=0;i<n;i++) { p = (Link *)malloc(sizeof(Link)); p->data = i+1; head->next = p; head = p; } head->next = NULL; return head; } void Print(Link *head) { Link *p = head->next; while(p!=NULL) { printf("%d\t",p->data); p = p->next; } } Link *Intersection(Link *La,Link *Lb,Link *Lc) { Link *pa,*pb,*pc,*p; pc = Lc; pa = La->next; pb = Lb->next; while(pa!=NULL&&pb!=NULL) { if(pa->data<pb->data) { pa = pa->next; } if(pa->data>pb->data) { pb = pb->next; } else { p = (Link *)malloc(sizeof(Link)); p->data = pa->data; pc->next = p; pc = p; pa = pa->next; pb = pb->next; } } pc->next = NULL; return Lc; } int main() { int a,b; Link *La,*Lb,*Lc; La = (Link *)malloc(sizeof(Link)); Lb = (Link *)malloc(sizeof(Link)); Lc = (Link *)malloc(sizeof(Link)); La->next = NULL; Lb->next = NULL; Lc->next = NULL; printf("请输入链表a的数目："); scanf("%d",&a); printf("\n请输入链表b的数目："); scanf("%d",&b); CreateLa(La,a); printf("链表la为：\n"); Print(La); CreateLb(Lb,b); printf("\n链表lb为：\n"); Print(Lb); Intersection(La,Lb,Lc); printf("\n链表lc为：\n"); Print(Lc); puts("\n"); system("pause"); }