[leetcode] Linked List Cycle II


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?

与上一题类似,这里我直接复用并修改了上一题的代码。

我们这样来想:

设置两个指针fast和slow,fast指针每次走2步,slow指针每次走1步。

假设进入环之前的路径长为n,环的长度为m。那么当slow指针进入环的第一个节点时,slow指针走了n步,fast指针在环里走了n步。两个指针的距离为m-n步。slow指针每走1步,fast指针都会追近1步。那么,当slow指针走了m-n步时,两个指针相遇。此时,fast-slow指针距离下次抵达环入口的距离为n步。

在fast-slow指针相遇时,在链表入口处放置一个指针p,p指针每次走一步。slow指针也每次走一步。这样当slow和p都再走n步时,两者在环的入口处相遇。即返回结果。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        ListNode * p = hasCycle(head);
        if(p == NULL){//no circle
            return NULL;
        }
        ListNode * q = head;
        while(q != p){
            p = p->next;
            q = q->next;
        }
        return p;
    }
    ListNode * hasCycle(ListNode *head) {
        ListNode * fast, * slow;
        if(head == NULL){
            return NULL;
        }
        fast = slow = head;
        while(true){
            if(fast->next == NULL || fast->next->next == NULL){
                return NULL;//no circle
            }
            fast = fast->next->next;
            slow = slow->next;
            if(fast == slow){
                return fast;
            }
        }
    }
};

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