142. Linked List Cycle II（js）

142. Linked List Cycle II

Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

Note: Do not modify the linked list.

 

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0
Output: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: no cycle
Explanation: There is no cycle in the linked list.

 

Follow up:
Can you solve it without using extra space?

题意：判断链表中是否存在环，如果在则找出入口节点

代码如下：

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */

/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var detectCycle = function(head) {
    if(!head || !head.next) return null;
    let fNode=head,sNode=head;
    while(fNode && fNode.next){
        fNode=fNode.next.next;
        sNode=sNode.next;    
        if(fNode == sNode) break;
    }
    //如果不是环
    if(fNode==null || fNode.next==null) return null;
    //是环，则sNode=head,fNode步长变为1，二者再次相遇在环的入口节点
    sNode=head;
    while(fNode!=sNode){
        fNode=fNode.next;
        sNode=sNode.next;
    }
    return sNode;
};