LeetCode 206 单链表翻转

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 = 1;
    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("
");
}

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 = 2; i < 10; i++){
        pushNode(head,i);
    }
    printListNode(head);
    head = reverseList(head);
    printListNode(head);
}