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

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 = None

class 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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