LeetCode :: Linked List Cycle I and II

I.Given a linked list, determine if it has a cycle in it.

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

Can you solve it without using extra space?

一、两道是连着一起出的，可以说第一道题目是做第二道题目的基础，第一题也是很经典的一个题目——判断一个链表是否包含闭环。

设两个指针，一快一慢，同时步进（慢的每次步进一、快的每次步进二），那么假如快的那个最终遇到NULL，就是说明没有闭环，如果快的那一个指针，最终和慢的相遇（如果不构成环形那么一快一慢永远不会相遇），说明有闭环。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    bool hasCycle(ListNode *head) {
        if (head == NULL)
            return false;
        ListNode *fast, *slow;
            fast = slow = head;
        while(fast != NULL && fast -> next != NULL){
            fast = fast -> next -> next;
            slow = slow -> next;
            if (fast == slow)
                return true;
        }
        return false;
    }
};

class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        ListNode *cur = head;
        ListNode *fast = hasCycle(head);
        if (!fast)
            return NULL;
        while(cur != fast){
            cur = cur -> next;
            fast = fast -> next;
        }
            return cur;
    }
private:
    ListNode *hasCycle(ListNode *head) {
        if (head == NULL)
            return NULL;
        ListNode *fast, *slow;
            fast = slow = head;
        while(fast != NULL && fast -> next != NULL){
            fast = fast -> next -> next;
            slow = slow -> next;
            if (fast == slow)
                return fast;
        }
        return NULL;
    }
};

LeetCode :: Linked List Cycle I and II,布布扣,bubuko.com

LeetCode :: Linked List Cycle I and II

原文：http://blog.csdn.net/u013195320/article/details/22521993