[C语言][LeetCode][61]Rotate List

题目

Rotate List
Given a list, rotate the list to the right by k places, where k is non-negative.

For example:
Given 1->2->3->4->5->NULL and k = 2,
return 4->5->1->2->3->NULL.

标签

Linked List、Two Pointers

难度

中等

分析

题目意思是按给定的value旋转单链表。我的做法是先获取链表的长度，然后从头遍历，找到新表的头结点，然后将新表结尾指向到旧表的头。

C代码实现

struct ListNode* rotateRight(struct ListNode* head, int k) {    struct ListNode *p, *t, *q, *ret;    int length = 0;    if(!head)        return NULL;    p = head;    while(p)    {        p = p->next;        length++;    }    p = head;    if(k>length)        k = length - k%length;    else        k = (length-k)%length;    while(p && k>1)    {        q = p;        p = p->next;        k--;    }    if(!p->next  || k == 0)        return head;    ret = t = p->next;    p->next = NULL;    while(t)    {        q = t;        t = t->next;    }    q->next = head;    return ret;}

这里看到别人写的一个更简单的，先找到旧表的尾节点，将尾节点的NEXT指向旧表的头，然后再去遍历找到新表的头结点

struct ListNode* rotateRight(struct ListNode* head, int k) {    int i;    if(head == NULL || k == 0) {        return head;    }    struct ListNode* p = head;    int len = 1;    while(p->next != NULL) {        p = p->next;        len++;    }    p->next = head;    k %= len;    for(i = 0; i < len - k; i++) {        p = p->next;    }    head = p->next;    p->next = NULL;    return head;}