哈哈，这题和环形链表1差不多 ，链表1详情参见 https://blog.csdn.net/a66666_/article/details/105211023
解法1. 快慢指针 （建议不要直接复制到leetcode，会报错，自己手打一遍）
时间复杂度O(N) 空间复杂度O(1)
java:
public ListNode detectCycle(ListNode head) {
ListNode walker = head;
ListNode runner = head;
while (runner != null && runner.next != null) {
walker = walker.next;
runner = runner.next.next;
if (walker == runner) {
ListNode walker2 = head;
while (walker != walker2) {
walker = walker.next;
walker2 = walker2.next;
}
return walker;
}
}
return null;
}
javsScript:
var detectCycle = function(head) {
let slower = head;
let flast = head;
while(flast != null && flast.next != null) {
slower = slower.next;
flast = flast.next.next;
if(slower === flast) {
let slower2 = head;
while(slower != slower2) {
slower = slower.next;
slower2 = slower2.next;
}
return slower;
}
}
return null;
};
相比 环形链表1的解法多了下方的代码
ListNode walker2 = head;
while (walker != walker2) {
walker = walker.next;
walker2 = walker2.next;
}
因为环形链表1快慢指针相遇的节点并不是连接的那个节点
如图所示，p为快慢指针相遇节点，q为连接节点，而连接节点由公式 a+2b+c == 2(a+b)可推出a == c
解法二：哈希表
时间复杂度O（N），空间复杂度O(N)
java描述：
public ListNode detectCycle(ListNode head) {
Set<ListNode> node = new HashSet<ListNode>();
while (head != null) {
if (node.contains(head)) {
return head;
}
node.add(head);
head = head.next;
}
return null;
}
javaScript描述：
var detectCycle = function(head) {
let nodes = new Set();
while(head) {
if (nodes.has(head)) return head;
nodes.add(head);
head = head.next;
}
return null;
};