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≤n1 \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

  1. // Linked List Cycle II
  2. // 时间复杂度O(n)，空间复杂度O(1)
  3. class Solution {
  4. public:
  5.     ListNode *detectCycle(ListNode *head) {
  6.         ListNode *slow = head, *fast = head;
  7.         while (fast && fast->next) {
  8.             slow = slow->next;
  9.             fast = fast->next->next;
  10.             if (slow == fast) {
  11.                 ListNode *slow2 = head;
  13.                 while (slow2 != slow) {
  14.                     slow2 = slow2->next;
  15.                     slow = slow->next;
  16.                 }
  17.                 return slow2;
  18.             }
  19.         }
  20.         return nullptr;
  21.     }
  22. };

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原文: https://soulmachine.gitbooks.io/algorithm-essentials/content/cpp/linear-list/linked-list/linked-list-cycle-ii.html