linear-list/linked-list/linked-list-cycle-ii

Linked List Cycle II

描述

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?

分析

当 fast 与 slow 相遇时，slow 肯定没有遍历完链表，而 fast 已经在环内循环了n圈($$1 \leq n$$)。假设 slow 走了s步，则 fast 走了2s步（fast 步数还等于s加上在环上多转的n圈），设环长为r，则：

2s = s + nr

s = nr

设整个链表长L，环入口点与相遇点距离为a，起点到环入口点的距离为x，则

x + a = nr = (n – 1)r +r = (n-1)r + L - x

x = (n-1)r + (L – x – a)

L – x – a为相遇点到环入口点的距离，由此可知，从链表头到环入口点等于n-1圈内环+相遇点到环入口点，于是我们可以从head开始另设一个指针slow2，两个慢指针每次前进一步，它俩一定会在环入口点相遇。

代码

// Linked List Cycle II
// 时间复杂度O(n)，空间复杂度O(1)
class Solution {
    public ListNode detectCycle(ListNode head) {
        ListNode slow = head, fast = head;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
            if (slow == fast) {
                ListNode slow2 = head;

                while (slow2 != slow) {
                    slow2 = slow2.next;
                    slow = slow.next;
                }
                return slow2;
            }
        }
        return null;
    }
}

// Linked List Cycle II
// 时间复杂度O(n)，空间复杂度O(1)
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) {
                ListNode *slow2 = head;

                while (slow2 != slow) {
                    slow2 = slow2->next;
                    slow = slow->next;
                }
                return slow2;
            }
        }
        return nullptr;
    }
};

相关题目

• Linked List Cycle
• Intersection of Two Linked Lists