160. 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.

解题思路：

（1）首先遍历两个链表得到他们的长度，就能知道哪个链表比较长，以及长的链表比短的链表多几个节点。

（2）在第二次遍历的时候，在较长的链表上先走若干步，接着同时在两个链表上遍历，找到第一个相同的结点就是他们的第一个公共结点。

/**
 * 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 or headB==NULL)return NULL;
        int length1=getLength(headA), length2=getLength(headB);
        int dist = abs(length1-length2);
        ListNode *Long_head =  headB, *Short_head = headA;
        if(length1>length2)
        {
            Long_head =  headA;
            Short_head = headB;
        }
        
        while(dist>0)
        {
            Long_head = Long_head->next;
            dist--;
        }
        while(Long_head!=NULL and Short_head!=NULL)
        {
            if(Long_head==Short_head)
                return Short_head;
            Long_head = Long_head->next; 
            Short_head = Short_head->next;
        }
        
        return NULL;
    }
    
    int getLength(ListNode *head)
    {
        if(head==NULL)return 0;
        int count = 1;
        while(head->next!=NULL)
        {
            count++;
            head = head->next;
        }
        return count;
    }
};