Given a linked list, return the node where the cycle begins. If there is no cycle, return null
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Follow up:
Can you solve it without using extra space?
/** * 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; //确定第一次相遇点 ListNode* p1=head; ListNode* p2=head; while(p2){ //p1向前移动一步 p1=p1->next; //p2向前移动两步 p2=p2->next; if(p2) p2=p2->next; if(p2 && p1==p2) break; } //如果没有环 if(p2==NULL)return NULL; //如果有环 //p2指向链表头 p2=head; //p1p2都已一步速向前推进,直至相遇 while(p1!=p2){ p1=p1->next; p2=p2->next; } return p1; } };
LeetCode: Linked List Cycle II [142],布布扣,bubuko.com
LeetCode: Linked List Cycle II [142]
原文地址:http://blog.csdn.net/harryhuang1990/article/details/35780221