CareerCup-2.5

Given a circular linked list, implement an algorithm which returns node at the beginning of the loop.

DEFINITION

Circular linked list: A (corrupt) linked list in which a node’s next pointer points to an earlier node, so as to make a loop in the linked list

EXAMPLE

input: A -> B -> C -> D -> E -> C [the same C as earlier]

output: C

我写了第一个方法,时间复杂度是Ο(n^2),答案是第二个方法,想了会才明白,复杂度是Ο(n)

#include <iostream>
using namespace std;

// Creating a Linked List:
struct Node
{
    Node(int d):data(d){next = NULL;};
    int data;
    Node* next;
};

Node* findPar(Node* head)
{
    if(head == NULL) return NULL;
    Node *p, *q = head->next;
    int countq = 0, countp = 0;
    while(q != head && q != NULL)
    {
        countq ++;
        p = head;
        countp = 0;
        while(p != q)
        {
            countp ++;
            p = p->next;
        }
        if(countp != countq)
            return p;
        else
            q = q->next;
    }
    if(q == head)
        return head;
    else
        return NULL;
};

/* 设head距离环交点k,环长度为n。q以2倍p的速度前进。p到达环交点的时候
 * q已走出环交点k步,q向前n-k步是p。继续前进当他们相遇时,p又走出n-k步,
 * 距离环交点n-k。q回到head,与p以相同速度前进,q走到环交点的时候p距离
 * 环交点n,他们此时在交点相遇。 
 */ 
Node* findPar2(Node* head)
{
    Node *p = head;
    Node *q = head;
    while(p->next != NULL)
    {
        p = p->next;
        q = q->next->next;
        if(p == q)
        {
            break;
        }
    }
    if(q->next == NULL)
    {
        return NULL;
    }
    p = head;
    while(p != q)
    {
        p = p->next;
        q = q->next;
    }
    return q;
};

int main()
{
    Node* head = new Node(1);
    head->next = new Node(2);
    head->next->next = new Node(3);
    head->next->next->next = new Node(4);
    head->next->next->next->next = new Node(5);
    head->next->next->next->next->next = head->next->next;
    cout<<findPar(head)->data<<endl;
    cout<<findPar2(head)->data<<endl;
    system("pause"); 
};

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