LeetCode 61. Rotate List
2017-06-22 17:33
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61. Rotate List
一、问题描述
Given a list, rotate the list to the right by k places, where k is non-negative.二、输入输出
For example:Given
1->2->3->4->5->NULLand k =
2,
return
4->5->1->2->3->NULL.
三、解题思路
这道题的意思是:把后面的k个数字移动到前面去。如果k比链表的长度要大,就需要取模运算。(吃了没文化的亏╮(╯▽╰)╭)Tow Pointers有人用fast slow指针来找末尾的那个位置,我还不知道怎么找。。
算法思路:遍历一遍链表,计算出链表的长度并把tail->next = head 形成一个循环链表。然后再从头开始往后遍历链表,走 length - k % length步,就找到新的链表尾部了,把链表断开就行了。想清楚了题目的意思其实并不难。
/** * 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* tail = head, *nTail = head; int length = 1; while(tail->next){ tail=tail->next; length++; } tail->next = head; int steps = length - k % length; while (--steps){ nTail = nTail->next; } head = nTail->next; nTail->next = NULL; return head; } };
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