https://leetcode.com/problems/reverse-linked-list/
思路很简单，分别设置三个结点，之后依次调整结点1和结点2的指向关系。
Before: pre -> nxt -> nxtnxt -> ..... Here current = pre,nxt = pre->next, nxtnxt = nxt->next.
After: pre <- nxt nxtnxt -> ..... Here current = nxt, nxt = nxtnxt
代码如下（附加了一个简单的初始化链表的实例，VS2012下测试PASS）
#include "stdio.h"
#include "stdlib.h"
#include "string.h"
#include <ctype.h> struct ListNode {
int val;
struct ListNode *next;
}; struct ListNode* initListNode(void){
struct ListNode* head = (struct ListNode*)malloc(sizeof(struct ListNode));
head->val = ;
head->next = NULL;
} void pushNode(struct ListNode* head, int val){
struct ListNode* tmp;
struct ListNode* pre = head;
while (pre->next != NULL)
pre = pre->next;
tmp = (struct ListNode*)malloc(sizeof(struct ListNode));
pre->next = tmp;
tmp->val = val;
tmp->next = NULL;
} void printListNode(struct ListNode* head){
struct ListNode* pre = head;
while (pre != NULL){
printf("%-4d", pre->val);
pre = pre->next;
}
printf("\n");
} struct ListNode* reverseList(struct ListNode* head) {
struct ListNode* current;
struct ListNode* nxt;
struct ListNode* nxtnxt; if (head == NULL)
return head;
else if (head->next == NULL)
return head;
else{
current = head;
nxt = current->next;
while (nxt != NULL){
nxtnxt = nxt->next;
nxt->next = current;
current = nxt;
nxt = nxtnxt;
}
head->next = NULL;
return current;
}
} int main(){
int i;
struct ListNode* head = initListNode();
for (i = ; i < ; i++){
pushNode(head,i);
}
printListNode(head);
head = reverseList(head);
printListNode(head);
}