LeetCode 61 — Rotate List（C++ Java Python）

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题目：http://oj.leetcode.com/problems/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.

题目翻译：

给定一个链表，向右旋转k个位置，其中k是非负的。
例如：
给定1->2->3->4->5->NULL和k = 2，
返回4->5->1->2->3->NULL。

分析：

把链表连成环，再在特定的位置断开。注意k可能大于链表的长度。

C++实现：

/** * 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 || k == 0)    {    return head;    }    int length = 1;    ListNode *node = head;    while(node->next != NULL) {    ++length;    node = node->next;    }    node->next = head;    int m = k % length;    for(int i = 0; i < length - m; ++i)    {    node = node->next;    }    head = node->next;    node->next = NULL;    return head;    }};

Java实现：

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public ListNode rotateRight(ListNode head, int n) {        if (head == null || n == 0) {return head;}int length = 1;ListNode node = head;while (node.next != null) {++length;node = node.next;}node.next = head;int m = n % length;for (int i = 0; i < length - m; ++i) {node = node.next;}head = node.next;node.next = null;return head;    }}

Python实现：

# Definition for singly-linked list.# class ListNode:#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution:    # @param head, a ListNode    # @param k, an integer    # @return a ListNode    def rotateRight(self, head, k):        if head == None or k == 0:            return head                length = 1        node = head        while node.next != None:            length += 1            node = node.next                    m = k % length                node.next = head                for i in range(length - m):            node = node.next                head = node.next                node.next = None                return head

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