day04之链表不带环相交+带环相交问题+fork函数输出几个短线

struct ListNode
{
int val;
ListNode *next;
ListNode(int x):val(x),next(NULL) {  }
};

bool IsCross(ListNode *phead1, ListNode *phead2)
{
if(phead1 == NULL || phead2 == NULL)
return false;

ListNode *p1 = phead1;
ListNode *p2 = phead2;

while(p1->next)
{
p1 = p1->next;
}
while(p2->next)
{
p2 = p2->next;
}
return p1==p2;
}

size_t ListSize(ListNode *phead)
{
if(phead == NULL)
return 0;

int size = 0;
while(phead)
{
size++;
phead = phead->next;
}
return size;
}

ListNode *GetCross(ListNode *phead1, ListNode *phead2)
{
if(phead1 == NULL || phead2 == NULL)
return NULL;

if(! IsCross(phead1, phead2))
return NULL;

ListNode *p1 = phead1;
ListNode *p2 = phead2;
int size1 = ListSize(phead1);
int size2 = ListSize(phead2);

int step = size1-size2;
if(step > 0)
{
while(step--)
{
p1 = p1->next;
}
while(p1 != p2)
{
p1 = p1->next;
p2 = p2->next;
}
}
else
{
step = -step;
while(step--)
{
p2 = p2->next;
}
while(p1 != p2)
{
p1 = p1->next;
p2 = p2->next;
}
}

return p1;
}

struct ListNode
{
int val;
ListNode *next;
ListNode(int x):val(x),next(NULL) {  }
};

ListNode *IsHaveLoop(ListNode *phead)
{
if(phead == NULL)
return NULL;

ListNode *pfast = phead;
ListNode *pslow = phead;

while(pfast && pfast->next)
{
pfast = pfast->next->next;
pslow = pslow->next;

if(pfast == pslow)
return pfast;
}

return NULL;
}

//求环的入度点
ListNode *EntryNode(ListNode *phead)
{
ListNode *pmeetnode = IsHaveLoop(phead);
if(pmeetnode == NULL)
return NULL;

ListNode *p1 = phead;
ListNode *p2 = pmeetnode;

while(p1 != p2)
{
p1 = p1->next;
p2 = p2->next;
}
return p1;
}

size_t ListSize(ListNode *phead)
{
if(phead == NULL)
return 0;

int size = 0;
while(phead)
{
size++;
phead = phead->next;
}
return size;
}

ListNode *noloop(ListNode *phead1, ListNode *phead2)
{
if(phead1 == NULL && phead2 == NULL)
return NULL;

ListNode *p1 = phead1;
ListNode *p2 = phead2;

while(p1->next)
{
p1 = p1->next;
}
while(p2->next)
{
p2 = p2->next;
}
if(p1 != p2)   //无环，不相交。
return NULL;
//无环，相交。
int size1 = ListSize(phead1);
int size2 = ListSize(phead2);
p1 = phead1;
p2 = phead2;
int step = size1-size2;
if(step > 0)
{
while(step--)
{
p1 = p1->next;
}
while(p1 != p2)
{
p1 = p1->next;
p2 = p2->next;
}
}
else
{
step = -step;
while(step--)
{
p2 = p2->next;
}
while(p1 != p2)
{
p1 = p1->next;
p2 = p2->next;
}
}
return p1;
}

ListNode *bothloop(ListNode *phead1, ListNode *phead2)
{
if(phead1 == NULL || phead2 == NULL)
return NULL;

ListNode *pEntryNode1 = EntryNode(phead1);
ListNode *pEntryNode2 = EntryNode(phead2);

if(pEntryNode1 == pEntryNode2) //相交第一种情况： 交点在入度点或入度点之前。
{
ListNode *p1 = phead1;
ListNode *p2 = phead2;
int size1 = 0;
int size2 = 0;
while(p1 != pEntryNode1)
{
size1++;
p1 = p1->next;
}
while(p2 != pEntryNode2)
{
size2++;
p2 = p2->next;
}

int step = size1-size2;
p1 = phead1;
p2 = phead2;

if(step > 0)
{
while(step--)
{
p1 = p1->next;
}
while(p1 != p2)
{
p1 = p1->next;
p2 = p2->next;
}
}
else
{
step = -step;
while(step--)
{
p2 = p2->next;
}
while(p1 != p2)
{
p1 = p1->next;
p2 = p2->next;
}
}

return p1;
}
else   //入度点不同，两种情况，一种情况是相交在环上，另一种是不相交
{
ListNode *pcur = pEntryNode1->next;
while(pcur != pEntryNode1)
{
if(pcur == pEntryNode2)
{
return pEntryNode2; //相交在环上
}
pcur = pcur->next;
}
}
return NULL;  //不相交
}

//此时链表可能带环，（1）两个链表不带环有可能相交，（2）一个带环一个不带环肯定不可能相交 （3）两个带环可能相交
ListNode *GetCross(ListNode *phead1, ListNode *phead2)
{
if(phead1 == NULL || phead2 == NULL)
return NULL;

ListNode *pmeetnode1 = IsHaveLoop(phead1);
ListNode *pmeetnode2 = IsHaveLoop(phead2);

if(pmeetnode1 == NULL && pmeetnode2 == NULL)// 情况一：都无环
{
return noloop(phead1, phead2);
}
else if(pmeetnode1 != NULL && pmeetnode2 != NULL) //情况二：都带环
{
return bothloop(phead1, phead2);
}
return NULL;  //情况三：一个带环，一个不带环，肯定不相交。
}

#include <stdio.h>
#include <unistd.h>
#include <sys/stats.h>

int main(void)
{
int i;
for(i=0; i<2; i++){
fork();
printf("-");
}

return 0;
}

最终输出8个短线，printf函数是行缓冲，当i = 0,第一个进程和第二个进程都有一个短线，因为没有\n，没有刷新，当i = 1,第三个和第四个进程都继承父进程一个短线，然后这四个进程再次输出短线，最终输出8个短线