141、142 Linked List Cycle & II

141、 Linked List Cycle

Given a linked list, determine if it has a cycle in it.

Follow up:

Can you solve it without using extra space?

思路

非常简单的一道题，如果链表有环，则在遍历的时候永远不会结束，一直在环内转圈。

所以，设置快慢指针，如果相遇，则说明有环。

代码（C++）

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
bool hasCycle(ListNode *head)
{
if (NULL == head || NULL == head->next)
return false;
ListNode* slow = head;
ListNode* fast = head;
while (fast->next && fast->next->next)
{
slow = slow->next;
fast = fast->next->next;
if (slow == fast)
return true;
}
return false;
}
};

142、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.

思路

在上一题的基础上，slow和fast相遇后，slow和head同时出发，每次一个步长，相遇的时候就是交点。

代码（C++）

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *detectCycle(ListNode *head)
{
if (NULL == head || NULL == head->next)
return false;
ListNode* slow = head;
ListNode* fast = head;
while (fast->next && fast->next->next)
{
slow = slow->next;
fast = fast->next->next;
if (slow == fast)
{
ListNode* tmp = head;
while (tmp != slow)
{
tmp = tmp->next;
slow = slow->next;
}
return tmp;
}
}
return NULL;
}
};