单链表的反转 - O(n)时间复杂度的非递归实现

　　其实这个才是逆序输出链表的正确操作，记得前面写的时候有四种办法，但除了递归比较简洁而且时间复杂度和空间复杂度都满足外，其他的方法都太丑陋了...

　　Code：

/* 反转链表的函数 */
struct reverseList * reverseLinkList(struct reverseList * node)
{
    struct reverseList * prev, * rear;
 
    rear = node -> next;
    while(rear -> next != NULL)            //整体并未向后移动
    {
        prev = rear -> next;            //先用一个容器指针来复制一份该指针
        rear -> next = prev -> next;    //把被复制的指针后面的一位赋给被复制的指针，意味着告诉被复制的指针的前一个指针，它的下一个是被复制指针的后一个。
        prev -> next = node -> next;    //把第一个指针赋给被复制指针的后一个，意味着告诉被复制的指针，它的下一个为第一个
        node -> next = prev;            //把复制的指针赋给第一个，意味着告诉头，它的下一个为复制指针
    }
 
    return node;
}

　　我推荐可以看看我参考别人的图文解释的文章：点击(ง •_•)ง

　　完整的代码测试：

/*
 *    反转单链表 - 非递归实现
 */
 
#include<stdio.h>
#include<stdlib.h>
 
struct reverseList {
    struct reverseList * next;
    int val;
};
 
struct reverseList * CreateList(void)
{
    struct reverseList * head, * s, * current;
    int i;
 
    head = (struct reverseList *)malloc(sizeof(struct reverseList));
    s = head;
    for(i = 0; i < 10; i++)
    {
        current = (struct reverseList *)malloc(sizeof(struct reverseList));
        current -> val = i + 1;
        s -> next = current;
        s = s -> next;
    }
    s -> next = NULL;
 
    return head;
}
 
struct reverseList * reverse_Link_List(struct reverseList * node)
{
    struct reverseList * prev, * rear;
 
    rear = node -> next;
    while(rear -> next != NULL)
    {
        prev = rear -> next;
        rear -> next = prev -> next;
        prev -> next = node -> next;
        node -> next = prev;
    }
 
    return node;
}
 
int main(void)
{
    struct reverseList * head, * p;
 
    head = CreateList();
 
    p = head -> next;
    while(p)
    {
        printf("%d ", p -> val);
        p = p -> next;
    }
    printf("\n");
 
    p = reverse_Link_List(head) -> next;
 
    while(p)
    {
        printf("%d ", p -> val);
        p = p -> next;
    }
    printf("\n");
 
    p = head;
    while(p)
    {
        head = p;
        p = p -> next;
        free(head);
    }
    p = NULL;
 
    return 0;
}