leetcode 141. circular linked list

Given a list, the list is determined whether a ring. (Do not use extra space)

Example: abcb

Ideas:

1. Method pointer speed

Set two hands, fast hands each take two steps slow pointer each time step, if it is a ring, then the pointer will soon overtake slower pointer until the pointer is equal to the fast and slow pointer when it returns true. Otherwise it returns false

var hasCycle = function (head) {
         // 1. Analyzing the absence of 
        IF (! head || head.next) {
             return  to false 
        } 
        // 2. double pointer loop 
        var FAST = head.next
         var SLOW = head
         // 2.1 fast pointer each take two steps, each step by slow pointer, if the pointer has been a fast value, it proved to be the ring until the two are equal, exit 
        the while (the fAST && fast.next) {
             IF (the fAST == sLOW ) {
                 return  to true 
            } 
            FAST = fast.next.next 
            SLOW =slow.next 
        } 
        //2.2 Otherwise, it is not annular 
        return  to false 
    };

2. I do not know what method

var hasCycle = function(head) {
        while(head != null){
            if(head == head.next){
                return true;
            }
            if(head.next != null){
                head.next = head.next.next;
            }
            head = head.next;
        }
        return false;
    };

This approach will destroy the list and is not recommended