[leetcode 160]Intersection of Two Linked Lists

Question:

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.

分析：
两个链表的长度差dist肯定在相同节点之前链表部分的差。将长链表先走 长度dist步，然后再同步走。

代码：

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
private:
int lengthOfList(ListNode *head){
if(head == NULL)
return 0;
int result = 0;
ListNode * node;
node = head;
while(node != NULL){
result++;
node = node->next;
}
return result;
}
public:
ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
int aLen,bLen;
if(headA == NULL || headB == NULL)
return NULL;
else{
aLen = lengthOfList(headA);
bLen = lengthOfList(headB);
int dist;
if(aLen < bLen){
dist = bLen - aLen;
while(dist){
headB = headB->next;
dist--;
}
}
if(aLen > bLen){
dist = aLen - bLen;
while(dist){
headA = headA->next;
dist--;
}
}
while(headB != headA){
headB = headB->next;
headA = headA->next;
}
return headB;
}
}
};