今天的算法题目是如何判断一个单链表有环

类似于追及问题创建两个指针，p2每一步走两个，p1每一步走一个

画的有点丑，如图，如果有环的情况下，最终他们会相遇
那么代码如下

#include <iostream>
using namespace std;
typedef struct LNode
{
    int data;
    struct LNode* next;
}LinkNode;

bool Is_Circle(LNode *head)
{
    LNode* p1 = head;
    LNode* p2 = head;
    while (p2 != NULL && p2->next != NULL)
    {
        p1 = p1->next;
        p2 = p2->next->next;
        if (p1 == p2)
            return true;
    }
    return false;
}
int main()
{
    LinkNode* L1,*L2,*L3,*L4,* L5, * L6, * L7;
    L1=(LinkNode*)malloc(sizeof(LinkNode));
    L2 = (LinkNode*)malloc(sizeof(LinkNode));
    L3 = (LinkNode*)malloc(sizeof(LinkNode));
    L4 = (LinkNode*)malloc(sizeof(LinkNode));
    L5 = (LinkNode*)malloc(sizeof(LinkNode));
    L6 = (LinkNode*)malloc(sizeof(LinkNode)); 
    L7 = (LinkNode*)malloc(sizeof(LinkNode));
    L1->data = 5;
    L2->data = 3;
    L3->data = 7;
    L4->data = 2;
    L5->data = 6;
    L6->data = 8;
    L7->data = 1;
    L1->next = L2;
    L2->next = L3;
    L3->next = L4;
    L4->next = L5;
    L5->next = L6;
    L6->next = L7;
    L7->next = L4;
    cout<<Is_Circle(L1);
    return 0;
}

问题延伸 1：如果链表有环，如何求出环的长度
解法，当两个指针首次相遇，证明有环，再次相遇就是环的长度，因为p2比p1多走了一倍路程

#include <iostream>
using namespace std;
typedef struct LNode
{
    int data;
    struct LNode* next;
}LinkNode;

int Is_Circle(LNode *head)
{
    int count=0;//用来记录步数
    LNode* First_encounter=head;
    LNode* Second_encounter= head;
    LNode* p1 = head;
    LNode* p2 = head;
   
    while (p2 != NULL && p2->next != NULL)
    {
        p1 = p1->next;
        p2 = p2->next->next;
        if (p1 == p2)
        { 
            First_encounter = p1;      
            Second_encounter = p2;
        }
        break;
    }
   
    while (First_encounter != NULL && First_encounter->next != NULL)
    {
        count++;
        First_encounter = First_encounter->next;
       Second_encounter = Second_encounter->next->next;
        if (First_encounter == Second_encounter)
            break;     
    }
           return  count;
}
int main()
{
    LinkNode* L1,*L2,*L3,*L4,* L5, * L6, * L7;
    L1=(LinkNode*)malloc(sizeof(LinkNode));
    L2 = (LinkNode*)malloc(sizeof(LinkNode));
    L3 = (LinkNode*)malloc(sizeof(LinkNode));
    L4 = (LinkNode*)malloc(sizeof(LinkNode));
    L5 = (LinkNode*)malloc(sizeof(LinkNode));
    L6 = (LinkNode*)malloc(sizeof(LinkNode)); 
    L7 = (LinkNode*)malloc(sizeof(LinkNode));
    L1->data = 5;
    L2->data = 3;
    L3->data = 7;
    L4->data = 2;
    L5->data = 6;
    L6->data = 8;
    L7->data = 1;
    L1->next = L2;
    L2->next = L3;
    L3->next = L4;
    L4->next = L5;
    L5->next = L6;
    L6->next = L7;
    L7->next = L4;
    cout<<Is_Circle(L1);
    return 0;
}

问题延伸 2 如何判断入环节点

指针p1一次只走一步，所走的距离是D+S1
指针p2一次只走二步，所走的距离是D+S1+n（s1+s2）
所以p2比p1的路程多走一倍
2(D+S1)=D+S1+n(S1+S2)
D=(n-1)(S1+S2)+S2
所以入环那段距离等于首次相遇点再走s2
此时把一个点放回开始处，两个点同时走一步，那么相遇点就是入环点了

#include <iostream>
using namespace std;
typedef struct LNode
{
    int data;
    struct LNode* next;
}LinkNode;

int Is_Circle(LNode *head)
{
    int count=0;//用来记录步数
 
    LNode* p1 = head;
    LNode* p2 = head;
   
    while (p2 != NULL && p2->next != NULL)
    {
        p1 = p1->next;
        p2 = p2->next->next;
        if (p1 == p2)
        break;
    }
    p1 = head;
    while (p1 != p2 && p1 != NULL)
    {
        p1 = p1->next;
        p2 = p2->next;
    }
    return p2->data;
}
int main()
{
    LinkNode* L1,*L2,*L3,*L4,* L5, * L6, * L7;
    L1=(LinkNode*)malloc(sizeof(LinkNode));
    L2 = (LinkNode*)malloc(sizeof(LinkNode));
    L3 = (LinkNode*)malloc(sizeof(LinkNode));
    L4 = (LinkNode*)malloc(sizeof(LinkNode));
    L5 = (LinkNode*)malloc(sizeof(LinkNode));
    L6 = (LinkNode*)malloc(sizeof(LinkNode)); 
    L7 = (LinkNode*)malloc(sizeof(LinkNode));
    L1->data = 5;
    L2->data = 3;
    L3->data = 7;
    L4->data = 2;
    L5->data = 6;
    L6->data = 8;
    L7->data = 1;
    L1->next = L2;
    L2->next = L3;
    L3->next = L4;
    L4->next = L5;
    L5->next = L6;
    L6->next = L7;
    L7->next = L4;
    cout<<Is_Circle(L1);
    return 0;
}