【Rotate List】cpp
题目:
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
.
代码:
/**
* 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 || !(head->next)) return head;
ListNode *p = head;
int len = ;
for (;p->next;++len,p=p->next){}
ListNode *end = p;
k = k % len;
p = head;
for (size_t i = ; i < len-k-; ++i)
{
p = p->next;
}
end->next = head;
head = p->next;
p->next = NULL;
return head;
}
};
Tips:
思路很简单,需要注意的是对指针的操作。
=====================================
第二次过这道题:
(1)没有考虑k比ListNodes长度大的情况
(2)没考虑k加上双指针的边界情况没有考虑完全,所以几次都没有AC。
完备的思路应该是:先求出来ListNodes的长度,k%len就是真正要rotate的元素。搞清楚这之后,再利用快慢指针常规思路就可以解出来了。
/**
* 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 || !head->next ) return head;
int len = ;
for ( ListNode* p = head; p->next; ++len, p=p->next){}
k = k % len;
ListNode dummpy(-);
dummpy.next = head;
ListNode* p1 = head;
ListNode* p2 = head;
for ( int i=; i<k; ++i ) p2 = p2->next;
while ( p2->next )
{
p1 = p1->next;
p2 = p2->next;
}
p2->next = head;
dummpy.next = p1->next;
p1->next = NULL;
return dummpy.next;
}
};
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