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该题目对内存的使用极其变态。所用变量不能超过4个。否侧会内存超限。解法有两个，其中第二种解法，内存还需要优化，否则会内存越界。
解法一：
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
ListNode* tempA = headA;ListNode* tempB = headB;
int countA=0,countB=0;
if(headA == NULL || headB == NULL)
return NULL;
while(tempA->next != NULL)
{
countA++;
tempA = tempA->next;
}
while(tempB->next != NULL)
{
countB++;
tempB = tempB->next;
}
if(tempA != tempB)
return NULL;
if(countA > countB)
{
int count = countA- countB;
while(count>0)
{
headA = headA->next;
count--;
}
}
else
{
int count = countB - countA;
while(count>0)
{
headB = headB->next;
count--;
}
}
while(headA != headB)
{
headA = headA->next;
headB = headB->next;
}
return headA;
}
};
解法二：
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
ListNode * endA;
if(headA == NULL || headB == NULL)
return NULL;
endA = getEnd(headA);
endA->next = headA;
bool ret ;
ListNode * pmeetpoint = NULL;
ret = RoundCircle(headB,&pmeetpoint);
if(ret == false)
return NULL;
ListNode *presult;
presult = GetEntry(headB,pmeetpoint);
endA->next = NULL;
return presult;
}
ListNode* GetEntry(ListNode *headA, ListNode *pmeetpoint)
{
while(headA != pmeetpoint)
{
headA = headA->next;
pmeetpoint = pmeetpoint->next;
}
return headA;
}
ListNode * getEnd(ListNode *headA)
{
while(headA->next != NULL)
{
headA= headA->next;
}
return headA;
}
bool RoundCircle(ListNode *headB,ListNode** ppmeetpoint)
{
ListNode * pslow,*pquick;
pslow = headB;
pquick = headB;
while(1)
{
if(pquick->next ==NULL)
return false;
pquick = pquick->next;
if(pquick->next ==NULL)
return false;
pquick = pquick->next;
pslow = pslow->next;
if(pslow == pquick)
{
*ppmeetpoint = pslow;
return true;
}
}
}
};
Intersection of Two Linked Lists
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原文地址：http://www.cnblogs.com/xgcode/p/4211916.html
