Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
Follow up:
Can you solve it without using extra space?
141. Linked List Cycle 的拓展，这题要返回环开始的节点，如果没有环返回null。
解法：双指针，还是用快慢两个指针，相遇时记下节点。参考：willduan的博客
public class Solution {
public ListNode detectCycle( ListNode head ) {
if( head == null || head.next == null ){
return null;
}
// 快指针fp和慢指针sp，
ListNode fp = head, sp = head;
while( fp != null && fp.next != null){
sp = sp.next;
fp = fp.next.next;
//此处应该用fp == sp ，而不能用fp.equals(sp) 因为链表为1 2 的时候容易
//抛出异常
if( fp == sp ){ //说明有环
break;
}
}
//System.out.println( fp.val + " "+ sp.val );
if( fp == null || fp.next == null ){
return null;
}
//说明有环，求环的起始节点
sp = head;
while( fp != sp ){
sp = sp.next;
fp = fp.next;
}
return sp;
}
} Python：
class ListNode:
def __init__(self, x):
self.val = x
self.next = None def __str__(self):
if self:
return "{}".format(self.val)
else:
return None class Solution:
# @param head, a ListNode
# @return a list node
def detectCycle(self, head):
fast, slow = head, head
while fast and fast.next:
fast, slow = fast.next.next, slow.next
if fast is slow:
fast = head
while fast is not slow:
fast, slow = fast.next, slow.next
return fast
return None
C++：
class Solution {
public:
ListNode *detectCycle(ListNode *head) {
ListNode *slow = head, *fast = head;
while (fast && fast->next) {
slow = slow->next;
fast = fast->next->next;
if (slow == fast) break;
}
if (!fast || !fast->next) return NULL;
slow = head;
while (slow != fast) {
slow = slow->next;
fast = fast->next;
}
return fast;
}
};
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[LeetCode] 141. Linked List Cycle 链表中的环