LeetCode | Swap Nodes in Pairs（将链表中的元素两两交换）

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Given a linked list, swap every two adjacent nodes and return its head.

For example,
Given 1->2->3->4, you should return the list as 2->1->4->3.

Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

题目就是将链表中的结点两两反转。

其实实现很简单，设置三个指针，q和m是要交换的结点指针，p是q的父节点，用来让父节点指针指向新的子节点m。

实现中要考虑输入head为空、只有一个结点这两种情况。然后只有两个结点的情况也要单独处理，因为这影响到了头结点指针的指向。我在实现时没有设置头结点，如果设置了头结点，就能将只有两个结点的情况，合并到后面的程序当中。

实现比较简单，代码如下：

#include <stdio.h>#include <stdlib.h>typedef struct LNode{    int data;    struct LNode *next;}LNode,*LinkList;void CreateList(LinkList *head,int n);void SwapNodes(LinkList *head);int IsTwo(LinkList head);void PrintList(LinkList head);void DestoryList(LinkList *head);int main(){    LinkList head = NULL;    int n;    while(scanf("%d",&n) != EOF){        CreateList(&head,n);        PrintList(head);        SwapNodes(&head);        PrintList(head);        DestoryList(&head);    }    return 0;}void CreateList(LinkList *head,int n){    int data;    for(int i = 0;i < n;++i){        LinkList p = (LinkList)malloc(sizeof(LNode));        if(!p){            printf("malloc error!\n");            exit(1);        }        scanf("%d",&data);        p->data = data;        p->next = (*head);        *head = p;    }}void SwapNodes(LinkList *head){    LinkList p,q,m;    if(*head == NULL){        printf("NULL!\n");        return ;    }    if((*head)->next == NULL)        return ;    p = (*head)->next;  //处理头指针情况，如果设置头结点，就不用这么麻烦了    q = p->next;    (*head)->next = q;    p->next = *head;    *head = p;    p = (*head)->next;    while(IsTwo(p)){        q = p->next;        m = q->next;        q->next = m->next;        m->next = q;        p->next = m;        p = q;    }}int IsTwo(LinkList head){    if(head->next == NULL || head->next->next == NULL)        return 0;    return 1;}void PrintList(LinkList head){    LinkList p = head;    while(p){        printf("%d ",p->data);        p = p->next;    }    printf("\n");}void DestoryList(LinkList *head){    if(*head == NULL)        return ;    LinkList p = *head;    *head = NULL;    while(p){        LinkList q = p->next;        free(p);        p = q;    }}

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