LeetCode OJ：Linked List Cycle II（循环链表II）

【LeetCode OJ：Linked List Cycle II（循环链表II）】的更多相关文章

[LeetCode OJ] Linked List Cycle II—Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

/** * 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 *p1, *p2; //p1和p2从链表的第一个节点出发,p1每次移动一…

[LeetCode]141. Linked List Cycle判断循环链表

快慢指针用来判断循环链表记住快慢指针有四种常用的应用场景: 1.找到有序链表的中点,快指针到头的时候,慢指针就是中点. 2.判断是不是循环链表,快慢指针相遇就是 3.找到循环链表的起点,以链表头结点开始的结点和以快慢指针相遇结点为开始的结点,这两个结点相遇的地方就是开始 4.判断两个链表是不是相交.将其中一个链表收尾相接,然后判断另外一个链表是不是循环链表. 有个博客写的很好: http://blog.csdn.net/willduan1/article/details/50938210…

【Leetcode】Linked List Cycle II

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up: Can you solve it without using extra space? 思路:由[Leetcode]Linked List Cycle可知.利用一快一慢两个指针可以推断出链表是否存在环路. 如果两个指针相遇之前slow走了s步,则fast走了2s步.而且fast已经在长…

[LeetCode] 142. Linked List Cycle II 链表中的环 II

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up:Can you solve it without using extra space? 141. Linked List Cycle 的拓展,这题要返回环开始的节点,如果没有环返回null. 解法:双指针,还是用快慢两个指针,相遇时记下节点.参考:willduan的博客 Java: pu…

[LeetCode] 141. Linked List Cycle 链表中的环

Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 给定一个链表,判断是否有环存在.Follow up: 不使用额外空间. 解法:双指针,一个慢指针每次走1步,一个快指针每次走2步的,如果有环的话,两个指针肯定会相遇. Java: public class Solution { public boolean hasCycle(Li…

Java for LeetCode 142 Linked List Cycle II

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up: Can you solve it without using extra space? 解题思路,本题和上题十分类似,但是需要观察出一个规律,参考LeetCode:Linked List Cycle II JAVA实现如下: public ListNode detectCycle(Li…