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.

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.

分析：因为其后面的元素都是一样的，先得出每个列表的长度，然后将长度长的往后移动直到其后面的长度和短的一样，然后逐渐判断和短的列表的元素是不是一样，不一样的话则都往后移一个。

时间复杂度：O（M）

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {        if(headA==null||headB==null)            return null;        int lena=0;        int lenb=0;        ListNode temp=headA;        ListNode temp1=headB;        while(temp!=null){            lena++;            temp=temp.next;        }         while(temp1!=null){            lenb++;            temp1=temp1.next;        }        if(lena<lenb){            temp1=headB;            int i=lenb-lena;            while(i>0){                temp1=temp1.next;                i--;            }            temp=headA;        }else{            temp=headA;            int i=lena-lenb;            while(i>0){                temp=temp.next;                i--;            }            temp1=headB;        }        while(temp!=temp1){                temp=temp.next;                temp1=temp1.next;        }        return temp;    }}