LeetCode:Linked List Cycle

题目：判断单链表是否有环，要求O(1)空间复杂度。题目链接

利用快慢指针，开始两个指针均指向链表head，然后循环移动，慢指针移动1步，快指针则移动2步，如果链表有环则两个指针一定会相遇。代码如下：

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

当然还可以用倒转链表的方法来判断是否有环，沿着链表遍历，每次遍历把指针倒转，如果有环最后一定会回到head节点。不过这样会破坏链表结构 ，需要恢复。

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

 

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