Given a linked list, determine if it has a cycle in it.
Follow up:
Can you solve it without using extra space?
给定一个链表,判断是否有环存在。Follow up: 不使用额外空间。
解法:双指针,一个慢指针每次走1步,一个快指针每次走2步的,如果有环的话,两个指针肯定会相遇。
Java:
public class Solution {
public boolean hasCycle(ListNode head) {
ListNode slow = head, fast = head;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
if (slow == fast) return true;
}
return false;
}
}
Python:
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
class Solution:
# @param head, a ListNode
# @return a boolean
def hasCycle(self, head):
fast, slow = head, head
while fast and fast.next:
fast, slow = fast.next.next, slow.next
if fast is slow:
return True
return False
C++:
class Solution {
public:
bool hasCycle(ListNode *head) {
ListNode *slow = head, *fast = head;
while (fast && fast->next) {
slow = slow->next;
fast = fast->next->next;
if (slow == fast) return true;
}
return false;
}
};
类似题目:
[LeetCode] 142. Linked List Cycle II 链表中的环 II