141. Linked List Cycle

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

To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.

Follow up:

Can you solve it using O(1) (i.e. constant) memory?

链表判断环问题, 传统解法把链表节点全部保存下来, 然后遍历看下一个节点是否已经保存过了. 时间复杂度O(N). 空间复杂度O(N).

如果不想用空间, 则用两个指针,一个走两步,一个走一步,有环的时候他们一定相遇. 时间复杂度比较随机, 但小于O(N).

例如链表的环外长度为6, 环的长度为10. 则一共需要走10步他们就能相遇. 链表总长度16

分析: 假设走N步, 那么相遇的时候快指针一定比慢指针多走一圈

2N - N = 环长度.  N=环长度.所以

/**
 * 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(NULL==head)
            return false;
        ListNode *fast=head, *low=head;
        while(fast->next&&fast->next->next)
        {
            low=low->next;
            fast=fast->next->next;
            if(fast==low)return true;
        }
        return false;
    }
};