Sunday, November 16, 2014

[LeetCode] Linked List Cycle I, II

Linked List Cycle I
Given a linked list, determine if it has a cycle in it.
Follow up:
Can you solve it without using extra space?
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?
思路:
能解第二题必能解第一题。这题涉及两个部分:
1. 检测linked list是否有环:用双指针追击法:
快指针fast一次走两步,慢指针slow一次走一步。如果有环,两指针必定在某一时间相遇。fast不会跳过slow。因为如果fast跳过slow,那么前一步它们必已经相遇。
2. 如果有环,如何判断环的起始点。
假设linked list从head到环起始点s的长度为L,环的周长为C(两节点之间的长度为之间link的数量)
当fast与slow第一次相遇的位置记为m1,并假设m1离开环起始点s距离X,由于fast走的总路程一定是slow的两倍:
(L + X)*2 = L + n*C + X => L = n*C - X
从m1出发,走n*C - X的路程将回到s,而这段路程正好等于head到s之间的路程!所以第一次相遇后,将slow移到head,然后slow/fast同时以一次走一步的速度前进,直到它们第二次相遇,便是s了。

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

总结:
这里用了do...while,否则由于开始slow = fast = head,循环马上会终止。
fast每次移动的时候要判断是否遇到尾部,如遇到则说明无环。而slow不需要判断。




1 comment:

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