leetcode160---Intersection of Two Linked Lists(交叉点)

问题描述：

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.

给两个链表，找出它们交集的那个节点，要求时间复杂度O(n)，空间复杂度O(1)。

问题求解：

如图所示，两个链表有可能长度不等，总长度差d=abs(lengthA-lengthB)即是两链表首节点距离交叉点的长度差。因此只要让长的链表先移动d个位置，然后两个链表再一起往前移动，当相等时，即是交叉点。
代码：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public://O(lengthA+lengthB)，空间复杂度O(1)    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {        ListNode *pa = headA;        ListNode *pb = headB;        int la=0, lb=0;        //(1)计算A和B的长度        while(pa)        {            pa = pa->next; la++;        }        while(pb)        {            pb = pb->next; lb++;        }        pa = headA;//将pa,pb回到起点        pb = headB;        //(2)将较长的链表先移动abs(la-lb)个距离        if(la > lb)        {            int n = la-lb;            while(n--) pa = pa->next;        }        else        {            int n = lb-la;            while(n--) pb = pb->next;        }        //(3)同时往前推进，相等时，就是交集的节点。        while(pa != pb)        {            pa = pa->next;            pb = pb->next;        }        return pa;    }};