题目描述：
Write a program to find the node at which the intersection of two singly linked lists begins.
For example, the following two linked lists:
A: a1 → a2
↘
c1 → c2 → c3
↗
B: b1 → b2 → b3
begin to intersect at node c1.
Notes:
- If the two linked lists have no intersection at all, return
null. - The linked lists must retain their original structure after the function returns.
- You may assume there are no cycles anywhere in the entire linked structure.
- Your code should preferably run in O(n) time and use only O(1) memory.
solution：
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
if (headA == NULL || headB == NULL)
return NULL;
int lenA = ;
int lenB = ;
ListNode *pA = headA;
while (pA != NULL)
{
++lenA;
pA = pA->next;
}
ListNode *pB = headB;
while (pB != NULL)
{
++lenB;
pB = pB->next;
}
if (pA != pB)
return NULL;
pA = headA;
pB = headB;
if (lenA > lenB)
{
int k = lenA - lenB;
while (k--)
{
pA = pA->next;
}
}
else
{
int k = lenB - lenA;
while (k--)
{
pB = pB->next;
}
}
while (pA != pB)
{
pA = pA->next;
pB = pB->next;
}
return pA;
}
上述解法来自《剑指offer》，还有一种基于Hashset的方法。
solution：
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
if (headA == NULL || headB == NULL)
return NULL;
unordered_set<ListNode*> st;
while (headA != NULL)
{
st.insert(headA);
headA = headA->next;
}
while (headB != NULL)
{
if (st.find(headB) != st.end())
return headB;
headB = headB->next;
}
return NULL;
}