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?

同Leetcode 141 Linked List Cycle

性质：distance from head to 环开始点 ＝＝ distance from 双指针相遇的点 to 环开始点

证明：

指针a速度为1 指针b速度为2

记head到环开始处的距离为k，环的周长为r，从环开始处顺方向到相遇点的距离为d

则a到相遇点走过的路程为 Sa = k + d, Sb＝ k + nr + d （n为b多绕的圈数）

取n= 1 时 Sb = k + r + d 同时要满足 Sa * 2 = Sb

则 d = r - k, Sa = r, Sb = 2r 满足条件

显然 不存在n>1 使得成立的情况

var detectCycle = function(head) {
    if(!head)
        return null
    var a = head
    var b = head
    while(b.next && b.next.next){
        b = b.next.next
        a = a.next
        if(a===b){
            a = head
            while(a !== b){
                a = a.next
                b = b.next
            }
            return a
        }
    } 
    return null
}