题目描述：
输入两个链表，找出它们的第一个公共结点。
解题思路一：
时间复杂度：$O(n)$,空间复杂度：$O(1)$.
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class {
public:
ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2) {
ListNode *ff;
ListNode *h1 = pHead1, *h2 = pHead2;
int count1 = 0, count2 = 0, count3 = 0;
while(h1)
{
count1++;
h1 = h1->next;
}
while(h2)
{
count2++;
h2 = h2->next;
}
if(count1 >= count2)
{
count3 = count1 - count2;
while(count3)
{
pHead1 = pHead1->next;
count3--;
}
}
if(count1 < count2)
{
count3 = count2 - count1;
while(count3)
{
pHead2 = pHead2->next;
count3--;
}
}
while(pHead1 && pHead2)
{
if(pHead1 == pHead2)
{
ff = pHead2;
break;
}
pHead1 = pHead1->next;
pHead2 = pHead2->next;
}
return ff;
}
};
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解题思路二：
时间复杂度：$O(n)$,空间复杂度：$O(1)$.
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class {
public:
ListNode* FindFirstCommonNode( ListNode* pHead1, ListNode* pHead2) {
ListNode *p1 = pHead1;
ListNode *p2 = pHead2;
while(p1!=p2)
{
p1 = (p1==NULL ? pHead2 : p1->next);
p2 = (p2==NULL ? pHead1 : p2->next);
}
return p1;
}
};
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