Linked List Cycle（链表成环）

判断链表中是否有环

来源：https://leetcode.com/problems/linked-list-cycle

Given a linked list, determine if it has a cycle in it.

一块一慢两个指针，如果有环，两个指针必定会在某个时刻相同且都不为空

Java

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public boolean hasCycle(ListNode head) {
        if(head == null || head.next == null) {
            return false;
        }
        ListNode low = head.next, fast = head.next.next;
        while(fast != null && fast.next != null && low != fast) {
            low = low.next;
            fast = fast.next.next;
        }
        if(low == fast && low != null) {
            return true;
        }
        return false;
    }
}

Python

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def hasCycle(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        if head == None or head.next == None:
            return False
        fast = head.next
        low = head
        while fast != None and fast.next != None:
            if fast == low:
                return True
            fast = fast.next.next
            low = low.next
        return False

找到链表中环的起点

来源：https://leetcode.com/problems/linked-list-cycle-ii

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

快慢两个指针相遇时，快指针从头开始一步遍历，慢指针从相遇节点一步遍历，下次相遇的结点就是环的入口节点

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode detectCycle(ListNode head) {
        if(head == null || head.next == null) {
            return null;
        }
        ListNode low = head.next, fast = head.next.next;
        while(fast != null && fast.next != null && low != fast) {
            low = low.next;
            fast = fast.next.next;
        }
        fast = head;
        while(low != null && low != fast) {
            low = low.next;
            fast = fast.next;
        }
        return low;
    }
}

Python

# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    def EntryNodeOfLoop(self, pHead):
        # write code here
        if pHead == None:
            return pHead
        fast = pHead
        slow = pHead
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            if fast == slow:
                fast = pHead
                while fast:
                    if fast == slow:
                        return fast
                    fast = fast.next
                    slow = slow.next
        return None