【经典算法】: 求链表中倒数第K个结点

链表中倒数第K个节点的求法

方法有两个：

1，先算出来链表中有多少个节点，假设为count个，然后顺序遍历

找到count-k个时就可以截至

2，先对整个链表反序，然后倒过来找第K个即可

下面给出第一种方法：

static void GetKNumber(struct node** head_ref,int k){
int count = 0,temp=0;
struct node* head = *head_ref;
struct node* head_trail = *head_ref;
while(head!=NULL){
count++;
head = head->next;
}
while(head_trail!=NULL){
temp++;
if(temp-1==count-k){
cout<<"the k data is"<<head_trail->data<<endl;
}
head_trail = head_trail->next;
}
}

上面是函数，然后下面是完整的程序：

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
using namespace std;
struct node{
int data;
struct node* next;
};
void push(struct node** head_ref,int new_data){
struct node* new_node = (struct node*) malloc(sizeof(struct node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}

static void reverse(struct node** head_ref){      //用于翻转，把所有的对应关系都反过来
struct node* prev = NULL;
struct node* current = *head_ref;
struct node* next;
while(current!=NULL){
next = current->next;

current->next = prev;     //倒过来指
prev = current;
current = next;  //用来将循环往后拖
}
*head_ref = prev;   //代表着最后一个
}

//链表中倒数第K个节点
static void GetKNumber(struct node** head_ref,int k){
int count = 0,temp=0;
struct node* head = *head_ref;
struct node* head_trail = *head_ref;
while(head!=NULL){
count++;
head = head->next;
}
while(head_trail!=NULL){
temp++;
if(temp-1==count-k){
cout<<"the k data is"<<head_trail->data<<endl;
}
head_trail = head_trail->next;
}
}

void printList(struct node *head)

{

struct node *temp = head;

while(temp != NULL)

{

printf("%d  ", temp->data);

temp = temp->next;

}

}
int main(){
struct node* head = NULL;
push(&head,20);
push(&head,4);
push(&head,15);
push(&head,85);
push(&head,60);
push(&head,100);
printList(head);
GetKNumber(&head,2);
reverse(&head);
cout<<endl<<"翻转后的结果"<<endl;
printList(head);
return 0;
}