链表逆序和链表翻转

来源：互联网发布：java final 变量编辑：程序博客网时间：2024/05/22 05:18

1. 链表逆序

void Reverse(LinkNode *head){LinkNode *p = head->next, *q;head->next = NULL;while(p != NULL){q = p->next;           //保存p的下一个节点p->next = head->next;  //将p插入到head之后（头插法）head->next = p;p = q;                 //p后移}}

2. 链表翻转。给出一个链表和一个数k，比如链表1→2→3→4→5→6，k=2，则翻转后2→1→4→3→6→5，若k=3，翻转后3→2→1→6→5→4，若k=4，翻转后4→3→2→1→6→5

void Reverse(LinkNode *head, int k){LinkNode *p = head, *q = head, *r;int count = 0;while(q->next != NULL){        r = q->next;        //保存q的下一个节点q->next = r->next;  //删除r节点r->next = p->next;  //将r插入到p之后（头插法）p->next = r;if(count == 0)      //只有第一次q才需要向后移（在纸上画一下就明白了）{q = q->next;}if(++ count == k)   //如果移动了k次，则"头结点"p后移k个{count = 0;p = q;}};}

3. 输出链表的倒数第n个节点

void PrintLastButN(LinkNode *head, int n){LinkNode *p = head->next, *q = p;for(int i = 1; p != NULL && i < n; i ++)  //将p后移n-1位{p = p->next;}while(p != NULL && p->next != NULL)  {q = q->next;                          //q后移直至p->next为空p = p->next;}printf("%d\n", q->data);                  //输出节点}

4. 单链表变形。如 1 2 3 4 5 变为 1 3 5 4 2；如1 2 3 4 变为 1 3 4 2

void Transfer(LinkNode *head){LinkNode *p = head, *q, *s = head; while(p->next != NULL){q = p->next;            //保存p的下一个节点if(q->data % 2)         //节点值为奇数{p->next = q->next;  //删除q节点q->next = s->next;  //将q插入到s之后（尾插法）s->next = q;s = q;              //s后移}p = q;                  //p后移}p = s->next;                //奇数遍历完，s指向最后一个奇数，p指向第一个偶数s->next = NULL;             //s->next置空while(p != NULL){q = p->next;            //保存p的下一个节点p->next = s->next;      //将p插入到s之后（头插法）s->next = p;p = q;                  //p后移}}

5. 链表排序

void Sort(LinkNode *head){LinkNode *p = head->next, *q, *r;int min, temp;while(p != NULL){q = p;                  //保存未排序部分第一个节点min = 0xFFFFFFFF;while(q != NULL){if(q->data < min) {min = q->data;  //保存未排序部分最小值r = q;          //保存最小值节点}q = q->next;        //q后移}temp = p->data;         //交换p和r节点值p->data = r->data;r->data = temp;p = p->next;            //p后移，排序部分又多了一位}}

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