LeetCode刷题笔记(链表):linked-list-cycle-ii
2018-01-15 21:03
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转载请注明作者和出处:http://blog.csdn.net/u011475210
代码地址:https://github.com/WordZzzz/Note/tree/master/LeetCode
刷题平台:https://www.nowcoder.com/ta/leetcode
题 库:Leetcode经典编程题
编 者:WordZzzz
题目描述
解题思路
C版代码实现
Follow up:
Can you solve it without using extra space?
有环的情况下, 求链表的入环节点:遍历链表,把每个元素指向下个链表的指针赋值为NULL,则循环要么在链表结尾停止,要么在环状链表入口处停止。
系列教程持续发布中,欢迎订阅、关注、收藏、评论、点赞哦~~( ̄▽ ̄~)~
完的汪(∪。∪)。。。zzz
代码地址:https://github.com/WordZzzz/Note/tree/master/LeetCode
刷题平台:https://www.nowcoder.com/ta/leetcode
题 库:Leetcode经典编程题
编 者:WordZzzz
题目描述
解题思路
C版代码实现
题目描述
Given a linked list, return the node where the cycle begins. If there is no cycle, returnnull.Follow up:
Can you solve it without using extra space?
解题思路
同linked-list-cycle-i一题,使用快慢指针方法,判定是否存在环,并记录两指针相遇位置(Z);有环的情况下, 求链表的入环节点:遍历链表,把每个元素指向下个链表的指针赋值为NULL,则循环要么在链表结尾停止,要么在环状链表入口处停止。
C++版代码实现
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: //判断是否为环状链表 bool hasCycle(ListNode *head) { if(head == NULL || head->next == NULL) return false; ListNode *fast = head; ListNode *slow = head; while(fast->next != NULL && fast->next->next != NULL){ slow = slow->next; fast = fast->next->next; if(slow->val == fast->val) return true; } return false; } //检测函数 ListNode *detectCycle(ListNode *head) { if(hasCycle(head)) { ListNode *temp = NULL; while(head ->next) { temp = head ->next; head ->next = NULL; head = temp; } return head; } else return NULL; } };
系列教程持续发布中,欢迎订阅、关注、收藏、评论、点赞哦~~( ̄▽ ̄~)~
完的汪(∪。∪)。。。zzz
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