LeetCode -- Linked List Cycle II

LeetCode -- Linked List Cycle II

題目描述：


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?




判斷鏈表是否有環，如果存在，返回環起始節點；如果不存在，返回Null。


思路：
1. 使用快慢指針的方法找到環的位置。
2. 如果找到了環，慢指針回到起點，快慢指針每次各走一步，下一次相遇的位置就是環的起點。




實現代碼：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode DetectCycle(ListNode head) 
    {
        if(head == null){
    		return null;
    	}


        var p = head;
    	var q = head;
    	
    	var found = false;
    	while(p != null && q != null && q.next != null && !found){
    		var t = q;
    		p = p.next;
    		q = q.next.next;
    		if(ReferenceEquals(p,q)){
    			found = true;
    		}
    	}
    	
        if(!found){
    		return null;
    	}
    	
    	// p start from head again
    	// and q standing where it is
    	// next time they meet point is where cycle starts from
    	p = head;
    	while(!ReferenceEquals(p, q)){
    		p = p.next;
    		q = q.next;
    	}
    	
    	return q;
    }
}