数据结构与算法 - 单链表逆转

数据结构与算法 - 单链表逆转

单链表逆转输出

方案一：head 作为已知首节点，最后节点指向null， 使用三个指针便利链表，逐个节点进行反转

实现思路图解：

实现代码：

struct ActList {
    ActList * next;
};

ActList * reverseList(ActList * head) {

    if (head == NULL || head -> next == NULL) {
        // 少于两个节点无需反转
        return head;
    }

    ActList * p;
    ActList * q;
    ActList * r;

    p = head;
    q = head -> next;
    head -> next = NULL;

    while (q) {
        r = q -> next;
        q -> next = p;

        p = q;
        q = r;
    }

    head = p;

    return head;
}

方案二： 对于一条链表，从第2个节点到第N个节点，依次逐节点插入到第1个节点(head节点)之后，(N-1)次这样的操作结束之后将第1个节点挪到新表的表尾

图示：

代码：

ActList* ReverseList2(ActList* head)
{
    ActList* p;
    ActList* q;
    p=head->next;
    while(p->next!=NULL){
        q=p->next;
        p->next=q->next;
        q->next=head->next;
        head->next=q;
    }

    p->next=head;//相当于成环
    head=p->next->next;//新head变为原head的next
    p->next->next=NULL;//断掉环
    return head;  
}

方案三：循环

ActList * reverseList3 (ActList * head) {
    if (head == NULL || head -> next == NULL) {
        // 少于两个节点无需反转
        return head;
    }

    ActList *pre = NULL;
    ActList *next = NULL;
    while (head != NULL) {
        next = head -> next;
        head -> next = pre;

        pre = head;
        head = next;
    }

    return pre;
}

方案四: 递归

ActList * reverseList4 (ActList * head) {
    if (head == NULL || head -> next == NULL) {
        // 少于两个节点无需反转
        return head;
    }
    ActList *newHead = reverseList4(head -> next);

    head -> next -> next = head;
    head -> next = NULL;

    return newHead;
}