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
.
一个小错误, 简直可以说是笔误, 浪费我近两个小时
悲夫~
启示:
当判断是否为NULL,时
如果是指针, 写成NULL == p 形式
如果是整形, 写成0 == n;
while(!np->next) 这里不是!, 去掉!
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: /*ListNode *kthNode(ListNode *head, int k) { if(!head||k==0||!head->next) return head; ListNode *np = head; ListNode *temp; int len = 1; while(!np->next) { len++; np = np->next; } np->next = head; k = k%len; k = len - k ; while(k--) np = np->next; temp = np->next; np->next = NULL; return temp; } */ ListNode *rotateRight(ListNode *head, int k) { if(!head||k==0||!head->next) return head; ListNode *np = head; ListNode *temp; int len = 1; while(np->next) { len++; np = np->next; } np->next = head; k = k%len; k = len - k ; while(k--) np = np->next; temp = np->next; np->next = NULL; return temp; } };