leetcode_Intersection of Two Linked Lists_easy_主要是方法

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.

方法：计算两个链表的长度差len,然后让长的先走len长度，后面2个链表同步走找重合的节点即可。

/** * 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=0,lenB=0;        ListNode *tmpA=headA,*tmpB=headB;        while(tmpA!=NULL)        {            lenA++;            tmpA=tmpA->next;        }        while(tmpB!=NULL)        {            lenB++;            tmpB=tmpB->next;        }        int len;        len=lenA>=lenB?(lenA-lenB):(lenB-lenA);        ListNode *h1,*h2;        h1=lenA>=lenB?headA:headB;//长的链表        h2=lenA<lenB?headA:headB;//短的链表        while(len-->0)            h1=h1->next;        while(h1!=h2)        {            h1=h1->next;            h2=h2->next;        }        if(h1!=NULL)            return h1;        else            return NULL;    }};