LeetCode 142. Linked List Cycle II

原题链接在这里：https://leetcode.com/problems/linked-list-cycle-ii/

题目：

Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

Note: Do not modify the linked list.

Follow up:
Can you solve it without using extra space? 

题解：

类似Linked List Cycle.

首先找到是否有cycle，若是没有返回null.

若是有cycle, 那么设a是head 到 cycle start 的距离，b是cycle start 到相遇点的距离，c是cycle的周长，s是walker现在走过的step数。

1. a + b + mc = s
2. a + b + nc = 2s

合并上面公式得到 a+b = (n-2m)*c. 也就是说a和b是关于 c 互补的, a+b一定是 c的倍数, 那么从现在的b点往前走a步一定是走了c的倍数回到cycle start点.

可以把walker放回head，runner从b点，两个pointer 都一次向前走一步，走过a步之后恰巧都走到了cycle start.

Time Complexity: O(n). Space: O(1).

AC Java:

/**
 * 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 walker = head;
        ListNode runner = head;
        while(runner.next != null && runner.next.next != null){
            walker = walker.next;
            runner = runner.next.next;
            if(walker == runner){
                walker = head;
                while(walker != runner){
                    walker = walker.next;
                    runner = runner.next;
                }
                return walker;
            }
        }
        return null;
    }
}

AC C++:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        if(!head || !head->next){
            return NULL;
        }

        ListNode *runner = head;
        ListNode *walker = head;
        while(runner && runner->next){
            runner = runner->next->next;
            walker = walker->next;
            if(runner == walker){
                walker = head;
                while(walker != runner){
                    runner = runner->next;
                    walker = walker->next;
                }

                return walker;
            }
        }

        return NULL;
    }
};

跟上Find the Duplicate Number.

参见了这篇帖子，是数学想法。