剑指offer之找到链表里面包含环的入口节点

1 问题

剑指offer之找到链表里面包含环的入口节点，比如

    //             node7<-node6 <-node5
    //              |              |
    //head->node1->node2->node3->node4

环的入口节点是node2

2 代码实现

#include <stdio.h>
#include <stdlib.h>

#define true 1
#define false 0 

typedef struct node
{
    int value;
    struct node *next;
} Node;

/**
 *得到环的第一个公共节点
 */
Node *getCommonNode(Node *head)
{
    if (head == NULL)
    {
        return NULL;
    }
    Node *first = NULL;
    Node *second = NULL;
    first = head;
    second = head;
int isCircle = false;
//判断是否有环
    while (second != NULL && (second->next) != NULL && (second->next->next != NULL))
    {
        first = first->next;
        second = second->next->next;
        if (first == second)
        {
            isCircle = true;
break;
        }
    }
if (isCircle == false)
{
printf("the list do not circle\n");
return NULL;
}
//判断环的大小,这个时候肯定是进到环里面去了
int len = 0;
first = first->next;
++len;
while (first != second)
{
len++;
first = first->next;
}

//求出入口节点
Node *start = head;
Node *end = head;
while (len-- > 0)
{
end = end->next;
}
while (start != end)
{
start = start->next;
end = end->next;
}
return start;
}

int main()
{
    Node *head = NULL;
    Node *node1 = NULL;
    Node *node2 = NULL;
    Node *node3 = NULL;
    Node *node4 = NULL;
    Node *node5 = NULL;
    Node *node6 = NULL;
    Node *node7 = NULL;
    head = (Node *)malloc(sizeof(Node));
    node1 = (Node *)malloc(sizeof(Node));
    node2 = (Node *)malloc(sizeof(Node));
    node3 = (Node *)malloc(sizeof(Node));
    node4 = (Node *)malloc(sizeof(Node));
    node5 = (Node *)malloc(sizeof(Node));
    node6 = (Node *)malloc(sizeof(Node));
    node7 = (Node *)malloc(sizeof(Node));
    if (head == NULL || node1 == NULL || node2 == NULL || node3 == NULL
        || node4 == NULL || node5 == NULL || node6 == NULL || node7 == NULL)
    {
        printf("malloc fail\n");
        return false;
    }
//             node7<-node6 <-node5
    //              |              |
    //head->node1->node2->node3->node4
    head->value = 0;
    head->next = node1;
    node1->value = 1;
    node1->next = node2;
    node2->value = 2;
    node2->next = node3;
    node3->value = 3;
    node3->next = node4;
    node4->value = 4;
    node4->next = node5;
    node5->value = 5;
    node5->next = node6;
    node6->value = 6;
    node6->next = node7;
    node7->value = 7;
    node7->next = node2;

    Node *result = getCommonNode(head);
    if (result != NULL)
    {
        printf("the first common value is %d\n", result->value);
    }
    else
    {
        printf("list do not have circle\n");
    }

    return true;
}

3 运行结果

the first common value is 2