Rotate List

文章目录

1. Rotate List

Rotate List

题目

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

思路

因为K是非负实数，可能超过链表长度，所以需要计算链表长度。遍历一遍链表，得到链表长度length，并且将首尾相连，形成一个循环链表。指针指到链表尾处，再移动length-k%length个节点，到达链表中断处p。按如下操作，就能达到rotate list效果。

p->next=head;
p->next=NULL;

解题

c++版

/**
 * 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 || k==0)
            return head;
        ListNode *p;
        p=head;
        int length=1;
        while(p->next){
            p=p->next;
            length++;
        }
        p->next=head;
        k=length-k%length;
        while(k){
            p=p->next;
            k--;
        }
        head=p->next;
        p->next=NULL;
        return head;
    }
};

python版

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    # @param head, a ListNode
    # @param k, an integer
    # @return a ListNode
    def rotateRight(self, head, k):
        if not head or k==0:
            return head
        length=1
        p=head
        while p.next:
            p=p.next
            length+=1
        p.next=head
        k=length-k%length
        while k>0:
            p=p.next
            k=k-1
        head=p.next
        p.next=None
        return head