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;    }};