Leetcode题目：Intersection of Two Linked Lists

题目：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. 

题目解答:

　　判断两个链表是否有交集，并返回相交的第一个节点，若不想交，返回NULL。

　　首先计算两个链表的长度，让长链表走到短链表头平行的位置之后，两个链表再一起开始走。并一边走，一边来判断是否节点是否相同。

代码如下：

/**
 * 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) {
        int lenA = getListLen(headA);
        int lenB = getListLen(headB);
        if((lenA == 0) || (lenB == 0))
            return NULL;
        ListNode *p = headA;
        ListNode *q = headB;
        if(lenA > lenB)
        {
            int i = lenA - lenB;
            while(i != 0)
            {
                i--;
                p = p -> next;
            }
        }
        else
        {
            int i = lenB - lenA;
            while(i != 0)
            {
                i--;
                q = q -> next;
            }
        }
        while((p != NULL) && (q != NULL) && (p != q) )
        {
            p = p -> next;
            q = q -> next;
        }
        if(p == q)
            return p;
        else
            return NULL;
    }
   
    int getListLen(ListNode *head)
    {
        int len = 0;
        ListNode *p = head;
        while(p != NULL)
        {
            len++;
            p = p-> next;
        }
        return len;
    }
};