环形链表 II

203. 环形链表 II

leetcode 链接

思路：

1. 先判断 head 以及 head.Next 是否有值，无则直接返回 null

2. 用 fast, slow 分别指向 head, fast 每次走两步，slow 每次走一步，当 fast === slow 时，跳出循环

3. 重新将 fast 指向 head, 然后 fast, slow 每次走一步，当 fast === slow 时，跳出循环

4. 返回 fast 或者 slow

typescript 解法

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function detectCycle(head: ListNode | null): ListNode | null {
  if (!head || !head.next) {
    return null;
  }
  let fast = head;
  let slow = head;
  while (fast && fast.next) {
    fast = fast.next.next;
    slow = slow.next;
    if (fast === slow) {
      break;
    }
  }
  fast = head;
  while (slow && slow != fast) {
    slow = slow.next;
    fast = fast.next;
  }
  return slow;
}

go 解法

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func detectCycle(head *ListNode) *ListNode {
    if head == nil || head.Next == nil {
        return nil
    }
    fast, slow := head, head


    for fast != nil && fast.Next != nil {
        slow = slow.Next
        fast = fast.Next.Next
        if slow == fast {
            break
        }
    }
    fast = head
    for slow != nil && fast != nil && fast != slow {
        slow = slow.Next
        fast = fast.Next
    }
    return slow
}