[Leetcode 141] Linked List Cycle

题目

判断单链表里是否有环

分析

用两个指针P1和P2，P1每次走一步，P2每次走两步，如果链表中有环，两个指针一定会在环中相遇。

C代码

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     struct ListNode *next; * }; */bool hasCycle(struct ListNode *head) {    if(head==NULL){        return false;    }    struct ListNode* quick = head;    struct ListNode* slow = head;    do{        if(quick==NULL || quick->next==NULL)            return false;        quick = quick->next->next;        slow = slow->next;    }while(quick!=slow);    return true;}

C++代码

/** * 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 || head->next==NULL){     //空指针和只有一个结点，输出false            return false;        }        ListNode* quick = head;        ListNode* slow = head;        while(quick!=NULL && quick->next!=NULL){  //注意循环结束条件，因为quick每次走两步            quick = quick->next->next;            slow = slow->next;            if(quick == slow){                return true;            }        }        return false;    }};