链表翻转(面试题目 C语言实现)

来源：互联网发布：python可以跨平台吗编辑：程序博客网时间：2024/04/27 21:23

题目：给出一个链表和一个数k，比如，链表为1→2→3→4→5→6，k=2，则翻转后2→1→6→5→4→3，若k=3，翻转后3→2→1→6→5→4，若k=4，翻转后4→3→2→1→6→5，用程序实现。

我的思路是：采用插入法翻转链表。

#include<stdio.h>#include<stdlib.h>typedef struct node{int data;struct node *next;};void CreateList(node *head,int length);void PrintList(node *head);void RotateList(node *head, int from, int to);void ShiftList(node *head, int k, int n);int main(){int len,k;node * head = (node*)malloc(sizeof(node));printf("input the list length:\n");scanf_s("%d", &len);CreateList(head, len);PrintList(head);printf("input k:\n");scanf_s("%d", &k);ShiftList(head, k, len);PrintList(head);system("pause");return 0;}void CreateList(node *head,int len){node *p,*q;head->data = NULL;p = (node*)malloc(sizeof(node));p = head;for (int i = 0; i < len; i++){q = (node*)malloc(sizeof(node));scanf_s("%d", &(q->data));q->next = NULL;p->next = q;p = p->next;}}void PrintList(node *head){node *temp = (node*)malloc(sizeof(node));temp = head;while (temp->next != NULL){temp = temp->next;printf("%d  ", temp->data);}printf("\n");}void RotateList(node *head, int from, int to){node *p, *q, *tail;p = (node*)malloc(sizeof(node));p = head;for (int i = 1; i < from; i++){p = p->next;}tail = (node*)malloc(sizeof(node));tail = p;p = p->next;q = (node*)malloc(sizeof(node));for (int i = from; i < to; i++){q = p->next;p->next = q->next;q->next = tail->next;tail->next = q;}}void ShiftList(node*head, int k,int n){RotateList(head, 1, k);RotateList(head, k+1, n);}

时间复杂度为O（n），空间复杂度为O（1）

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