leetcode 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.

Follow up:
Can you solve it without using extra space?

题目要求不修改链表结构，同时不用额外的内存，去找到链表中环的起点。

我们可以通过快慢指针很容易判断有没有环，接下来，再把快指针的步伐也变为1并从head开始走，那么快慢指针再次相遇对应的指针就是环的起点。

解释：可以把问题抽象成一个直线和圈的结构，

粉色点表示相遇点，那么快指针走过a+b+c+b,慢指针走过a+b. 又2(a+b)=a+b+c+b,所以c=a.

/**
 * 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) {
        ListNode* fast = head;
        ListNode* slow = head;
        if (head == nullptr) return head;
        while (fast->next != nullptr && fast->next->next != nullptr) {
            fast = fast->next->next;
            slow = slow->next;
            if (fast == slow) break;
        }
        if (fast->next == nullptr || fast->next->next == nullptr) return nullptr; 
        fast = head;
        while (fast != nullptr && slow != nullptr) {
            if (fast == slow) return fast;
            fast = fast->next;
            slow = slow->next;
        }
        return nullptr;
    }
};