Leetcode 141. Linked List Cycle & 142. Linked List Cycle II

Leetcode 141. Linked List Cycle

解法一

直接给遍历过的点打标记，再回来就说明有环。

class Solution {
public:
    const int N = 100005;
    bool hasCycle(ListNode *head) {
        ListNode *cur = head;
        while(cur) {
            if(cur->val == N) {
                return true;
            }
            cur->val = N;
            cur = cur->next;
        }
        return false;
    }
};

解法二

快慢指针法，跑得快的一定会在某个时刻追上跑得慢的，详细的时间复杂度分析：
官方知乎解答

class Solution {
public:
    const int N = 100005;
    bool hasCycle(ListNode *head) {
        ListNode *fast = head, *slow = head;
        while(fast && fast->next) {
            fast = fast->next->next;
            slow = slow->next;
            if(fast == slow) {
                return true;
            }
        }
        return false;
    }
};

注意要把if(fast == slow)的判定放在循环结束，否则一上来就会无脑return true.

142. Linked List Cycle II

用上面的方法一可以秒杀，用快慢指针就比较麻烦了。
这里有详细的数学推导
知乎优质解答

ListNode *detectCycle_1(ListNode *head) {
        ListNode *fast = head, *slow = head;
        bool hasCycle = false;
        while (fast && fast->next) {
            fast = fast->next->next;
            slow = slow->next;
            if (fast == slow) {
                hasCycle = true;
                break;
            }
        }
        if (!hasCycle) {
            return nullptr;
        }
		ListNode* newStart = slow;	//	新的起点
		slow = head;
		while(newStart != slow) {
			slow = slow->next;
			newStart = newStart->next;
		}
		return slow;
    }