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Given a linked list, determine if it has a cycle in it.
Follow up:
Can you solve it without using extra space?
算法
思路1：
快慢指针，当两个指针相遇，则表示有环，否则则无环，可能需要多遍遍历，空间复杂度O(1)，时间复杂度O(n)
1 public class Solution { 2 public boolean hasCycle(ListNode head) { 3 if(head == null) return false; 4 ListNode fast = head; 5 ListNode slow = head; 6 while(true){ 7 if(fast.next == null || fast.next.next == null) return false; 8 fast = fast.next.next; 9 slow = slow.next; 10 if(fast == slow) return true; 11 } 12 } 13 }
思路2：
hash表，当某个节点有两个前驱，则该节点是环的起点，而且只需一次遍历，空间复杂度O(n)，时间复杂度O(n)
1 public class Solution { 2 public boolean hasCycle(ListNode head) { 3 if(head == null) return false; 4 ListNode tem = new ListNode(0); 5 tem.next = head; 6 Map<ListNode,ListNode> hash = new HashMap<ListNode,ListNode>(); 7 while(true){ 8 if(tem.next == null) return false; 9 if(hash.get(tem.next) != null) return true; 10 hash.put(tem.next,tem); 11 tem = tem.next; 12 } 13 } 14 }
思路２同样是Linked List Cycle II的解法
ListNode的结构如下：
1 public class ListNode { 2 int val; 3 ListNode next; 4 ListNode(int x){ 5 val = x; 6 next = null; 7 } 8 }
[leetcode]Linked List Cycle,布布扣,bubuko.com
标签：style blog http color os art
原文地址：http://www.cnblogs.com/huntfor/p/3854969.html
