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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
.
题意:
翻转链表。
思路:
首先需要读懂题意。题目的描述有问题,应该是将链表的最后k个元素移动到链表的头部。
这道题的本质就是寻找链表的倒数第k个元素。在该点将链表分成两个部分,然后调换顺序即可。
陷阱:
k的长度没有说明,可能k比链表的长度还要大。 设链表的长度为Len,那么移动的节点个数应该为 K%Len.
class Solution { public: int getListLength(ListNode *head){ int len=0; ListNode *tmp=head; while(tmp){ len++; tmp = tmp->next; } return len; } ListNode *rotateRight(ListNode *head, int x) { if(head==NULL) return head; // empty list int len = getListLength(head); x = x%len; // how many nodes to rotate if( len<=1 || x==0) return head; // find xth node from tail of list ListNode *tmp=head; ListNode *lend=head; ListNode *hstart=head; while(tmp->next){ if(x==0){ lend = lend->next; }else{ --x; } tmp = tmp->next; } hstart = lend->next; lend->next=NULL; tmp->next=head; return hstart; } };
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[LeetCode 题解]: Rotate List,布布扣,bubuko.com
标签:des style blog http color strong io for
原文地址:http://www.cnblogs.com/double-win/p/3883882.html