[LintCode] 帶環連結串列 II Linked List Cycle II
阿新 • • 發佈:2019-01-24
給定一個連結串列,如果連結串列中存在環,則返回到連結串列中環的起始節點的值,如果沒有環,返回null。
樣例
給出 -21->10->4->5, tail connects to node index 1,返回10
挑戰
不使用額外的空間
Given a linked list, return the node where the cycle begins.
If there is no cycle, return null.
Example
Given -21->10->4->5, tail connects to node index 1,return 10
Challenge
Follow up:
Can you solve it without using extra space?
/**
* Definition for ListNode.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int val) {
* this.val = val;
* this.next = null;
* }
* }
*/
public class Solution {
/**
* @param head: The first node of linked list.
* @return: The node where the cycle begins.
* if there is no cycle, return null
*/
public ListNode detectCycle(ListNode head) {
if(null == head || head.next == null || head.next.next == null) return null;
ListNode fast = head, slow = head;
while(true) {
if(slow == null || fast == null || fast.next == null) {
return null;
}
fast = fast.next.next;
slow = slow.next;
if (fast == slow) {
break;
}
}
slow = head;
while(true) {
fast = fast.next;
slow = slow.next;
if(fast == slow) {
return fast;
}
}
}
}