160. 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.

我一开始的思路是用暴力的方法：将A中每一个结点的值都与B中每个结点的值进行比较，当相同时返回。提交结果是超时了，看了一下其他人的solution，发现有一个很好的方法：定义两个指针分别只想链表A和B的首节点。求出两链表的长度作差。举个例子：比如题目所给的case，lengthA=5，lengthB=6，则作差得到 lengthB- lengthA=1，将指针pb从链表B的首节点开始走1步，即指向了第二个节点，pa指向链表A首节点，然后它们同时走，每次都走一步，当它们相等时，就是交集的节点。

AC：

/** * 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) {        ListNode *pa=headA,*pb=headB;int lengthA=0,lengthB=0;if(pa==NULL||pb==NULL)return NULL;while(pa) {pa=pa->next;lengthA++;}while(pb) {pb=pb->next;lengthB++;}if(lengthA<=lengthB){int n=lengthB-lengthA;pa=headA;pb=headB;while(n) {pb=pb->next;n--;}}else{int n=lengthA-lengthB;pa=headA;pb=headB;while(n) {pa=pa->next;n--;}}while(pa!=pb){pa=pa->next;pb=pb->next;}return pa;    }};