142. Linked List Cycle II

Problem

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

Note: Do not modify the linked list. Follow up: Can you solve it without using extra space?

Solution

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

/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var detectCycle = function(head) {
  var slow = head;
  var fast = head;
  var entry = head;
  while (slow && fast) {
    slow = slow.next;
    fast = fast.next ? fast.next.next : undefined;
    if (slow === fast) {
      while (entry !== slow) {
        entry = entry.next;
        slow = slow.next;
      }
      return entry;
    }
  }
  return null;
};

Explain:

fast 每次走两步，slow 每次走一步，如果链表有环的话，它们最终会相遇。

设：

• x 为 head 到环起点的长度
• y 为环的长度
• a 为环起点到 fast 和 slow 相遇点的长度
• b 为 fast 和 slow 相遇点到环起点的长度

则有：

• fast 走的长度为 x + a
• slow 走的长度为 x + y + a
• (x + y + a) / (x + a) = 2

推出：y = x + a; x = y - a = b;

Complexity:

• Time complexity : O(n).
• Space complexity : O(1).