【剑指Offer】面试题23：链表中环的入口节点-优快云博客

本文链接：https://blog.youkuaiyun.com/DJames23/article/details/115148922

本文介绍了一种Java方法，用于在给定链表中判断是否存在环，并在环存在时找到环的入口节点，利用快慢指针技巧，O(1)空间复杂度实现。

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/*
 public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}
*/
public class Solution {

    public ListNode EntryNodeOfLoop(ListNode pHead) {
         //①判断是否包含环
        if (pHead == null){
            return null;
        }
        ListNode slow = pHead;
        ListNode fast = pHead;
        boolean flag = false;
        while (fast != null && fast.next != null){
            slow = slow.next;
            fast = fast.next.next;
            if (slow == fast){
                flag = true;
                break;
            }
        }
        if (!flag){
            //flag仍为false,说明不包含环
            return null;
        }else {
            //②记录环内节点的个数
            int count = 1;
            fast = fast.next;
            while(fast != slow){
                count++;
                fast = fast.next;
            }
            //③找到入口节点
            fast = pHead;
            slow = pHead;
            for (int i = 0; i < count; i++) {
                fast = fast.next;
            }
            while (fast != slow){
                fast = fast.next;
                slow = slow.next;
            }
            return fast;
        }
    }
}

public class Solution {
 
    public ListNode EntryNodeOfLoop(ListNode pHead)
    {
        if(pHead == null || pHead.next == null){
            return null;
        }
 
        ListNode fast = pHead;
        ListNode slow = pHead;
 
        while(fast != null && fast.next != null){
            fast = fast.next.next;
            slow = slow.next;
            if(fast == slow){
                ListNode slow2 = pHead;
                while(slow2 != slow){
                    slow2 = slow2.next;
                    slow = slow.next;
                }
                return slow2;
            }
        }
        return null;
 
    }
}