每间隔K个元素逆转一次单链表

例如：
Inputs:   1->2->3->4->5->6->7->8->9->NULL and k = 3

Output:   3->2->1->4->5->6->9->8->7->NULL. 

方法1：

#include<stdio.h>#include<stdlib.h> /* Link list node */struct node{    int data;    struct node* next;}; /* Reverses alternate k nodes and   returns the pointer to the new head node */struct node *kAltReverse(struct node *head, int k){    struct node* current = head;    struct node* next;    struct node* prev = NULL;    int count = 0;        /*1) reverse first k nodes of the linked list */    while (current != NULL && count < k)    {       next  = current->next;       current->next = prev;       prev = current;       current = next;       count++;    }       /* 2) Now head points to the kth node.  So change next        of head to (k+1)th node*/    if(head != NULL)      head->next = current;        /* 3) We do not want to reverse next k nodes. So move the current         pointer to skip next k nodes */    count = 0;    while(count < k-1 && current != NULL )    {      current = current->next;      count++;    }     /* 4) Recursively call for the list starting from current->next.       And make rest of the list as next of first node */    if(current !=  NULL)       current->next = kAltReverse(current->next, k);      /* 5) prev is new head of the input list */    return prev;} /* UTILITY FUNCTIONS *//* Function to push a node */void push(struct node** head_ref, int new_data){    /* allocate node */    struct node* new_node =            (struct node*) malloc(sizeof(struct node));     /* put in the data  */    new_node->data  = new_data;     /* link the old list off the new node */    new_node->next = (*head_ref);         /* move the head to point to the new node */    (*head_ref)    = new_node;} /* Function to print linked list */void printList(struct node *node){    int count = 0;    while(node != NULL)    {        printf("%d  ", node->data);        node = node->next;        count++;    }}     /* Drier program to test above function*/int main(void){    /* Start with the empty list */    struct node* head = NULL;    int i;    // create a list 1->2->3->4->5...... ->20    for( i= 20; i > 0; i--)      push(&head, i);      printf("Given linked list \n");     printList(head);      head = kAltReverse(head, 3);      printf("\nModified Linked list \n");     printList(head);      getchar();     return(0);}

时间复杂度为Ο(n)

方法2：

// 单链表.cpp : Defines the entry point for the console application.//#include "stdafx.h"#include <stdio.h>#include <stdlib.h> /* Link list node */struct node{    int data;    struct node* next;}; /* Helper function for kAltReverse() */struct node * _kAltReverse(struct node *node, int k, bool b); /* Alternatively reverses the given linked list in groups of    given size k. */struct node *kAltReverse(struct node *head, int k){  return _kAltReverse(head, k, true);}  /*  Helper function for kAltReverse().  It reverses k nodes of the list only if    the third parameter b is passed as true, otherwise moves the pointer k     nodes ahead and recursively calls iteself  */struct node * _kAltReverse(struct node *node, int k, bool b){   if(node == NULL)       return NULL;    int count = 1;   struct node *prev = NULL;   struct node  *current = node;   struct node *next;     /* The loop serves two purposes      1) If b is true, then it reverses the k nodes       2) If b is false, then it moves the current pointer */   while(current != NULL && count <= k)   {       next = current->next;        /* Reverse the nodes only if b is true*/       if(b == true)          current->next = prev;                    prev = current;       current = next;       count++;   }       /* 3) If b is true, then node is the kth node.        So attach rest of the list after node.      4) After attaching, return the new head */   if(b == true)   {        node->next = _kAltReverse(current,k,!b);        return prev;           }       /* If b is not true, then attach rest of the list after prev.      So attach rest of the list after prev */      else   {        prev->next = _kAltReverse(current, k, !b);        return node;          }}   /* Function to push a node */void push(struct node** head_ref, int new_data){    /* allocate node */    struct node* new_node = (struct node*) malloc(sizeof(struct node));                /* put in the data  */    new_node->data  = new_data;      /* link the old list off the new node */    new_node->next = (*head_ref);      /* move the head to point to the new node */    (*head_ref)    = new_node;}  /* Function to print linked list */void printList(struct node *node){    int count = 0;    while(node != NULL)    {        printf("%d  ", node->data);        node = node->next;        count++;    }}  /* Drier program to test above function*/int main(void){    /* Start with the empty list */    struct node* head = NULL;    int i;      // create a list 1->2->3->4->5...... ->20    for(i = 20; i > 0; i--)      push(&head, i);      printf("Given linked list \n");    printList(head);    head = kAltReverse(head, 3);      printf("\nModified Linked list \n");    printList(head);      getchar();    return(0);}

时间复杂度为Ο(n)