160. Intersection of Two Linked Lists

问题：

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.

答案：

•    将A，B两个链表看做两部分，交叉前与交叉后
•    交叉后的长度是一样的，因此交叉前的长度差即为总的长度差
•    只要去除这些长度差，距离交叉点就等距了
•    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 l1=1;
       int l2=1;
       int i;
       ListNode *p1=headA, *p2=headB;
       while(p1->next!=NULL){
           p1=p1->next;
           l1++;
       }
       while(p2->next!=NULL){
           p2=p2->next;
           l2++;
       }
       if(p1->val!=p2->val)
       return NULL;
       if(l1>l2){
           for(i=0;i<l1-l2;i++){
               headA=headA->next;
           }
       }
       if(l1<l2){
           for(i=0;i<l2-l1;i++){
               headB=headB->next;
           }
       }
       while(headA!=headB){
           headA=headA->next;
           headB=headB->next;
       }
       return headA; 
    }
};