[LeetCode]142. Linked List Cycle II

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

从第一次相遇可得：slow: a + b; fast: 2(a + b). 并且2a + 2b = a + b + n(b + c), 可得a = (n - 1)b + nc，以此slow移到head，然后slow和fast都每次移动一位，最后相遇位置就是环开头。

/** * Definition for singly-linked list. * class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public ListNode detectCycle(ListNode head) {        if (head == null || head.next == null) {            return null;        }        ListNode fast = head.next.next;        ListNode slow = head.next;        while (fast != null && fast.next != null) {            if (fast == slow) {                break;            }            fast = fast.next.next;            slow = slow.next;        }        if (fast == null || fast.next == null) {            return null;        }        slow = head;        // 这里fast也是每次移一位！        while (fast != slow) {            slow = slow.next;            fast = fast.next;        }        return slow;    }}