Intersection of Two Linked Lists

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.

解题思路：
1、可以将A,B两个链表看做两部分，交叉前与交叉后。
2、交叉后的长度是一样的，因此交叉前的长度差即为总长度差。
3、只要去除这些长度差，距离交叉点就等距了。
4、为了节省计算，在计算链表长度的时候，顺便比较一下两个链表的尾节点是否一样，若不一样，则不可能相交，直接可以返回NULL。

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
if(headA == NULL || headB == NULL)
return NULL;
int lenA = 1;
int lenB = 1;
ListNode *tempA = headA;
ListNode *tempB = headB;

while(tempA->next != NULL){
tempA = tempA->next;
lenA++;
}
while(tempB->next != NULL){
tempB = tempB->next;
lenB++;
}

if(tempA != tempB)
return NULL;
if(lenA>lenB){
for(int i=0;i<lenA-lenB;i++){
headA = headA->next;
}
}
if(lenB>lenA){
for(int i=0;i<lenB-lenA;i++){
headB = headB->next;
}
}
while(headA != headB){
headA = headA->next;
headB = headB->next;
}
return headA;
}
};