对于一个给定的链表，返回环的入口节点，如果没有环，返回null 拓展： 你能给出不利用额外空间的解法么？
加载中...
import java.util.*; /** * Definition for singly-linked list. * class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { public ListNode detectCycle(ListNode head) { } }
/** * 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) { } };
# class ListNode: # def __init__(self, x): # self.val = x # self.next = None # # # @param head ListNode类 # @return ListNode类 # class Solution: def detectCycle(self , head ): # write code here
/* * function ListNode(x){ * this.val = x; * this.next = null; * } */ /** * * @param head ListNode类 * @return ListNode类 */ function detectCycle( head ) { // write code here } module.exports = { detectCycle : detectCycle };
# class ListNode: # def __init__(self, x): # self.val = x # self.next = None # # # @param head ListNode类 # @return ListNode类 # class Solution: def detectCycle(self , head ): # write code here
package main import . "nc_tools" /* * type ListNode struct{ * Val int * Next *ListNode * } */ /** * * @param head ListNode类 * @return ListNode类 */ func detectCycle( head *ListNode ) *ListNode { // write code here }
/** * struct ListNode { * int val; * struct ListNode *next; * }; */ /** * * @param head ListNode类 * @return ListNode类 */ struct ListNode* detectCycle(struct ListNode* head ) { // write code here }