判断单链表是否是循环链表以及找出循环链表入口

#include <iostream>
#include <vector>
#include <stack>
#include <queue>

using namespace std;

struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution {
public:
vector<int> postorderTraversal(TreeNode *root) {
vector<int> res;
if (!root){
return res;
}

stack<TreeNode*> s;

int sign = 1;
TreeNode* p = NULL;

do{
while (root){
//printf("val :%d in stack\n", root->val);
s.push(root);
root = root->left;
}

p = NULL;
sign = 1;
while (!s.empty() && sign){
root = s.top();

if (root->right == p){
res.push_back(root->val);
p = root;
s.pop();
}
else{
sign = 0;
root = root->right;
}
}
} while (!s.empty());

return res;
}

vector<int> preorderTraversal(TreeNode *root) {
vector<int> res;
if (!root){
return res;
}

stack<TreeNode*> s;
bool flag = false;

do{
while (root){
s.push(root);
root = root->left;
}

flag = false;

while (!s.empty()){
root = s.top();
s.pop();

printf("%d ", root->val);

if (root->right)
{
root = root->right;
flag = true;

break;
}

}
} while (!s.empty() || flag);

printf("\n");

return res;
}

vector<int> midorderTraversal(TreeNode *root) {
vector<int> res;
if (!root){
return res;
}

queue<TreeNode*> s;

do{
if (!s.empty())
{
root = s.front();
s.pop();
}

if(root){
printf("%d ", root->val);

if (root->left)
{
s.push(root->left);
}
if (root->right)
{
s.push(root->right);
}
}

} while (!s.empty());

printf("\n");

return res;
}
};

class node
{
public:
char c;
char d;

double i;

short s;

}no;

class A

{

int i;

char c1;

};

class B:public A

{

char c2;

};

class C:public B

{

char c3;

};

struct LinkList{
int val;
LinkList* next;
LinkList(int val):val(val){
next = NULL;
}
};
typedef LinkList ListNode;

void doseq(LinkList* node){
if (!node)
{
return;
}

LinkList* head = node;

LinkList* slow = head;
LinkList* quick = head;

while (quick)
{
quick = quick->next;

if (quick)
{
quick = quick->next;
slow = slow->next;
}
}

LinkList* cur = NULL;
stack<LinkList*> q;
LinkList* next = slow->next;
slow->next = NULL;
cur = next;
while (cur)
{
q.push(cur);
next = cur->next;
cur->next = NULL;
cur = next;
}

cur = head;

LinkList* cucur = NULL;
while (!q.empty() && cur)
{
next = q.top();
q.pop();

cucur = cur->next;

next->next = cur->next;
cur->next = next;

cur = cucur;
}
}

LinkList* insert(LinkList* node, int val){
while(node->next)
{
node = node->next;
}

LinkList* newnode = new LinkList(val);
node->next = newnode;

return newnode;
}

void print(LinkList* q)
{
while (q)
{
printf("%d ", q->val);
q = q->next;
}

}

bool hasCycle(ListNode *head) {
if (!head)
{
return false;
}

ListNode* slow = head;
ListNode* quick = head;

while (quick)
{
quick = quick->next;

if (quick)
{
quick = quick->next;
slow = slow->next;

if (quick == slow)
{
break;
}
}
}

slow = head;
while(slow != quick)
{
slow = slow->next;
quick = quick->next;
}

//printf("cycle begin node:%d\n", slow->val);

return false;
}
int main(){
LinkList* q1 = new LinkList(1);
insert(q1, 2);
insert(q1, 3);
insert(q1, 4);
LinkList* q2 = insert(q1, 5);
q2->next = q1;

printf("%res:%d\n", hasCycle(q1));
}

判断循环链表很简单，就是快慢两个指针；

但是如何找循环链表入口就需要分析下， 看下面的图：



假设quick,slow指针在Z点相遇，则slow走了a+b;quick走了a+b+c+b；

因为quick速度是slow的2倍，有(a+b)/t * 2 = (a+2b+c);

也就是a=c

这个结论很给力，此时qucik、slow在Z点，而我们现在想到循环入口Y点，头结点到Y点需要a步，而a=c，z再到y是c，所以此时从x,z点走当二者相遇就是Y点了。

就是这样。