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;
    }
};