LeetCode 61. Rotate List

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本文介绍了一种链表右旋转算法的实现方法，通过调整链表节点指针，将链表向右旋转k个位置。文章详细展示了如何通过一次遍历获取链表长度，并根据k值调整链表结构。

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;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* rotateRight(ListNode* head, int k) {
        if (head == NULL) {
            return head;
        }
        ListNode* newHead = head;
        ListNode* cur = head;
        ListNode* last;
        int n = 0;
        while(cur != NULL) {
            last = cur;
            cur = cur->next;
            ++n;
        }
        k = k % n;
        if (k == 0) {
            return head;
        }
        ListNode* pre;
        for (int i = n; i > k; --i) {
            pre = newHead;
            newHead = newHead->next;
        }
        pre->next = NULL;
        last->next = head;
        return newHead;
    }
};