061 - Rotate List

原创于 2015-12-30 16:33:53 发布 · 303 阅读

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本文介绍了一种算法，用于将链表向右旋转k个位置。通过遍历链表确定其长度，并利用模运算来减少不必要的旋转次数。文章详细解释了如何通过双指针法找到新的头结点和尾结点，最终实现链表的有效旋转。

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.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* rotateRight(struct ListNode* head, int k) 
{
	struct ListNode *first, *second, *newhead;
	int i;

	if (!head) return head;
	first = second = head;

	int total = 0;
	while (first) {
		first = first->next;
		total++;
	}
	k %= total;
	i = k;
	first = head;
	while (second && i--) second = second->next;
	if (!k || !second) return head;
	while (second->next) {
		first = first->next;
		second = second->next;
	}

	newhead = first->next;
	first->next = NULL;
	second->next = head;
	head = newhead;
	return newhead;
}