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 nullvar a = headvar b = headwhile(b.next && b.next.next){b = b.next.nexta = a.nextif(a===b){a = headwhile(a !== b){a = a.nextb = b.next}return a}} return null
}

转载于:https://www.cnblogs.com/lilixu/p/4589910.html

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