【LeetCode】141. Linked List Cycle 解題報告

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141. Linked List Cycle / Easy

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.

Example 1:

Input: head = [3,2,0,-4], pos = 1
IOutput: true
IExplanation: There is a cycle in the linked list, where the tail connects to the 1st node (0-indexed).

Example 2:

Input: head = [1,2], pos = 0
Output: true
Explanation: There is a cycle in the linked list, where the tail connects to the 0th node.

Example 3:

Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.

Constraints:

  • The number of the nodes in the list is in the range [0, 10^4].
  • -10^5 <= Node.val <= 10^5
  • pos is -1 or a valid index in the linked-list.

Solution: Floyd’s Cycle Finding Algorithm

思路

只要碰到在 Linked List 中判斷 Cycle 的題目，就少不了龜兔賽跑演算法（Floyd’s Cycle Finding Algorithm）
概念非常簡單，即有兩個指標，一個快一個慢。快指標一次走兩步，慢指標一次一步。
如果快指標走到底，代表沒有迴圈，但若快指標走到最後又和慢指標相會，代表有迴圈存在。

效能

Complexity

  • Time Complexity: O(N)
  • Space Complexity: O(1)

LeetCode Result

Code

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