Leetcode - 141 - Linked List Cycle

最新推荐文章于 2022-06-05 15:21:14 发布

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本文介绍两种有效的方法来判断链表中是否存在循环：使用哈希表记录遍历过的节点（时间复杂度O(n)，空间复杂度O(n)）；采用快慢指针技巧（时间复杂度O(n)，空间复杂度O(1)）。通过Java、Python和C++实现。

Given head, the head of a linked list, determine if the linked list has a cycle in it.

There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail's next pointer is connected to. Note that pos is not passed as a parameter.

Return true if there is a cycle in the linked list. Otherwise, return false.

Solutions:

(1) Hash table: save the nodes that have already been seen.

Time: O(n); Space: O(n)

Java:

public boolean hasCycle(ListNode head) {
    Set<ListNode> nodesSeen = new HashSet<>();
    while (head != null) {
        if (nodesSeen.contains(head)) {
            return true;
        } else {
            nodesSeen.add(head);
        }
        head = head.next;
    }
    return false;
}

// https://leetcode.com/problems/linked-list-cycle/solution/

(2) Two Pointers: Fast & slow pointers must meet in a loop

Java:

public class Solution {
    public boolean hasCycle(ListNode head) {
        if (head == null) {
            return false;
        }

        ListNode slow = head;
        ListNode fast = head.next;
        while (slow != fast) {
            if (fast == null || fast.next == null) {
                return false;
            }
            slow = slow.next;
            fast = fast.next.next;
        }
        return true;
    }
}
// https://leetcode.com/problems/linked-list-cycle/solution/


public boolean hasCycle(ListNode head) {
    if(head==null) return false;
    ListNode walker = head;
    ListNode runner = head;
    while(runner.next!=null && runner.next.next!=null) {
        walker = walker.next;
        runner = runner.next.next;
        if(walker==runner) return true;
    }
    return false;
}
// https://leetcode.com/problems/linked-list-cycle/discuss/44489/O(1)-Space-Solution


public class Solution {
    public boolean hasCycle(ListNode head) {
        ListNode walker = head;
        ListNode runner = head;
        while(runner!=null && runner.next!=null) {
            walker = walker.next;
            runner = runner.next.next;
            if(walker==runner) return true;
        }
        return false;
    }
}

Python:

def hasCycle(self, head):
    try:
        slow = head
        fast = head.next
        while slow is not fast:
            slow = slow.next
            fast = fast.next.next
        return True
    except:
        return False

# https://leetcode.com/problems/linked-list-cycle/discuss/44494/Except-ionally-fast-Python

C++:

class Solution {
public:
    bool hasCycle(ListNode *head) {
        if(head == NULL || head -> next == NULL)    
            return false;
 
        ListNode *fast = head;
        ListNode *slow = head;
        
        while(fast -> next && fast -> next -> next){
            fast = fast -> next -> next;
            slow = slow -> next;
            if(fast == slow)
                return true;
        }
 
        return false;
    }
};

//https://leetcode.com/problems/linked-list-cycle/discuss/44604/My-faster-and-slower-runner-solution


class Solution {
public:
bool hasCycle(ListNode *head) {
    ListNode* slow = head;
    ListNode* fast = head;
    
    while(fast && fast->next) {
        slow = slow->next;
        fast = fast->next->next;

        if(slow == fast) {
            return true;
        }
    }
    
    return false;
}
};
// with no need to check