題目描述
題目連結
解法
- 利用快慢指針 fastt 和 slow,fast 一次走兩步,slow 一次走一步
- 兩指針相遇說明有環,第一次相遇時所走步數的內插補點正好是環的倍數,但是是多少倍不可知
- 讓 fast從頭再走,slow 留在原地,fast 和 slow 均一次走一步,當兩個指針第二次相遇時,均處在環的起點處,傳回slow
/**
* 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(head == NULL || head->next == NULL)
return NULL;
ListNode *fast = head, *slow = head;
while(fast != NULL && fast->next != NULL)
{
slow = slow->next;
fast = fast->next->next;
if(slow == fast)
{
fast = head;
while(fast != slow)
{
fast = fast->next;
slow = slow->next;
}
return slow;
}
}
return NULL;
}
};
快慢指針的思路在 【二分查找】LeetCode - 287. 尋找重複數 的解法2中小試牛刀過