题目链接

注意点

k可能会大于链表长度

解法

解法一：首先遍历一遍链表得到链表的长度，k对其取余数。然后设置快慢指针，快指针先走k步，然后快慢指针一起走，当快指针走到底的时候，慢指针也走到了新的链表头结点的前一个结点，这时候修改快慢指针的指向就好了。时间复杂度O(n)。

/**
 * 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;
        int i,n = 0;
        ListNode* p = head;
        while(p != NULL)
        {
            p = p ->next;
            n++;
        }
        //cout << n;
        k %= n;
        ListNode* fast = head;
        ListNode* slow = head;
        while(k > 0)
        {
            fast = fast->next;
            k--;
        }
        //if(fast == NULL) return head;
        while(fast->next != NULL)
        {
            fast = fast->next;
            slow = slow->next;
        }
        fast->next = head;
        fast = slow->next;
        slow->next = NULL;
        return fast;
    }
};

解法二：同样首先遍历一遍链表得到链表的长度，k对其取余数。将链表尾和链表头相连，然后继续往后遍历（也就是从头开始）n-k个结点，这时正好指向新的链表头结点的前一个结点。时间复杂度O(n)

/**
 * 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;
        int i,n = 1;
        ListNode* p = head;
        while(p->next != NULL)
        {
            p = p ->next;
            n++;
        }
        p->next = head;
        k %= n;
        for(i = 0;i < n-k;i++) p = p->next;
        ListNode* ret = p->next;
        p->next = NULL;
        return ret;
    }
};

小结

链表题