[Leetcode] Linked List Cycle II

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?

转自:http://www.cnblogs.com/x1957/p/3406448.html

比I麻烦点的就是找到循环开始点TAT

I只是判断是否循环。要求不使用额外空间（不然hash就可以了

按I的思路，我们又慢指针S和快指针F。。。F走两步，S走一步。。。若有环，必定相遇。

画个图（很丑勿喷

假设在红色凸起的地方相遇了。

F走的路程应该是S的两倍

S = x + y

F = x + y + z + y = x + 2y + z

2*S = F

2x+2y = x + 2y + z

得到x = z

 

也就是从head到环开始的路程 ＝ 从相遇到环开始的路程

那么。。。只要S和F相遇了，我们拿一个从头开始走，一个从相遇的地方开始走

两个都走一步，那么再次相遇必定是环的开始节点！

代码也很精简，值得学习一下。

/**
 * 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) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if(head == NULL) return NULL;
        ListNode* S = head;
        ListNode* F = head;
        
        while(F != NULL){
            if(F) F = F -> next;
            if(F) F = F -> next;
            if(S) S = S -> next;
            if(F != NULL && F == S){
                S = head;
                while(S != F){
                    S = S -> next;
                    F = F -> next;
                }
                return S;
            }
        }
        return NULL;
    }
};

下面是自己写的：

/**
 * 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 == NULL) return NULL;
        ListNode *slow = head, *fast = head;
        while (slow != NULL && fast != NULL) {
            if (slow->next != NULL) {
                slow = slow->next;
            } else {
                return NULL;
            }
            if (fast->next != NULL) {
                fast = fast->next;
            } else {
                return NULL;
            }
            if (fast->next != NULL) {
                fast = fast->next;
            } else {
                return NULL;
            }
            if (slow == fast) {
                break;
            }
        }
        slow = head;
        while (slow != fast) {
            slow = slow->next;
            fast = fast->next;
        }
        return slow;
    }
};