题目描述:
不用辅助空间判断,链表中是否有环
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ /*一个指针走的快 一个指针走得慢*/ class Solution { public: bool hasCycle(ListNode *head) { if(head == NULL) return false; ListNode *first = head; ListNode *Second = head; while(Second->next != NULL){ first = first->next; Second = Second->next->next; if(Second == NULL) return false; if(Second == first) return true; } return false; } };
142 找到环开始的结点
第一次相遇时slow走过的距离:a+b,fast走过的距离:a+b+c+b。
因为fast的速度是slow的两倍,所以fast走的距离是slow的两倍,有 2(a+b) = a+b+c+b,可以得到a=c(这个结论很重要!)。
/** * 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) return NULL; ListNode *first = head; ListNode *second = head; bool flag = true; while(second->next != NULL){ first = first->next; second = second->next->next; if(second == NULL) return NULL; if(first == second){ flag = false; break; } } if(flag == true) return NULL; first = head; while(first != second){ first = first->next; second = second->next; } return first; } };
时间: 2024-10-10 07:02:21