只调整指针来交换两个相邻的元素【双链表】

n为链表长度，s为指定的位置，即将第s个数与第s+1个数对换

本次不再默认链表元素为1,2...n，而是手动输入n个元素

思路：在单链表对换后，把prior指针做调整就ok

#include<stdio h="">#include<stdlib h="">#include<string h="">typedef struct node{int data;struct node *next;struct node *prior;}Node;Node* InitList(int n){Node *l,*p,*s;int i,j;l=(Node*)malloc(sizeof(Node));l->next=NULL;p=(Node*)malloc(sizeof(Node));p=l;for(i=0;i<n i="" s="(Node*)malloc(sizeof(Node));" scanf="" d="" j="" s-="">data=j;s->next=p->next;p->next=s;s->prior=p;p=s;}return l;} //初始化 void PrintList(Node *l){Node *p;p=l->next;while(p){printf("%3d",p->data );p=p->next;}printf("\n");} //打印链表元素 void fun(Node *l,int s){Node *p,*tmp1,*tmp2;int i;p=l;for(i=0;i<s-1 i="" p="p-">next;}tmp1=p->next ;tmp2=p->next->next ;tmp1->next=tmp2->next;p->next=tmp2;tmp2->next=tmp1; //先把next指针换好tmp2->prior=p;tmp1->prior=tmp2;tmp1->next->prior=tmp1; //再逐个调整好prior指针 }int main(){Node *l;int n,s;printf("n=");scanf("%d",&n);printf("s=");scanf("%d",&s);l=InitList(n);printf("the original list:");PrintList(l);fun(l,s);printf("the new list:");PrintList(l);}</s-1></n></string></stdlib></stdio>