[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;
}