Two pointers (1) -- Linked List Cycle II, Rotate List

Linked List Cycle II

Given a linked list, return the node where the cycle begins. If there is no cycle, return 

null

.

Note: Do not modify the linked list.

Follow up:

Can you solve it without using extra space?
此题之前已有过介绍。 可以看这篇的解释，很详细：点击打开链接

ListNode *detectCycle(ListNode *head) {
if (head == NULL) return NULL;
ListNode* ptr = head;
ListNode* fastPtr = ptr;
do {
if (fastPtr == NULL || fastPtr->next == NULL)
return NULL;
ptr = ptr -> next;
fastPtr = fastPtr -> next -> next;
} while (ptr != fastPtr);
ptr = head;
while (ptr != fastPtr){
ptr = ptr -> next;
fastPtr = fastPtr -> next;
}
return ptr;
}

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

.
1. 先找出list的长度，然后找到断开处的指针。

2. 使用快慢指针的话并不能减少指针移动的操作，所以不用了

ListNode* rotateRight(ListNode* head, int k) {
if (head == NULL) return NULL;
int len = 1;
ListNode* ptr = head;
while (ptr -> next != NULL){
ptr = ptr -> next;
len++;
}
ListNode* tail = ptr;
k = k % len;
if (k == 0) return head;
ptr = head;
for (int i = 1; i < len - k; i++){
ptr = ptr -> next;
}
ListNode* newH = ptr -> next;
ptr -> next = NULL;
tail -> next = head;
return newH;
}