142. 环形链表 II

思路

• 快慢指针相遇，即为有环
• 从head，和相遇点开始，一步步的next，再次相遇点即为环的起始点

/**
 * 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;
        bool flag = false;
        // 快慢指针相遇即是有环
        while (fast && fast->next) {
            fast = fast->next->next;
            slow = slow->next;
            if (slow == fast) {
                flag = true;
                break;
            }
        }
        // 若有环
        if (flag == true) {
            int pos = 1;
            ListNode *reStart = head;
            ListNode *crossover = slow;
            while (reStart != crossover) {
                reStart = reStart->next;
                crossover = crossover->next;
                pos++;
            }
            return reStart;
        }

        return NULL;
        
    }
};

问题

• 空指针问题
• 超时因为我从head和slow->next开始的遍历，导致无法相遇。
• 总耗时16分钟