【LeetCode从零单排】No.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.

Credits:
Special thanks to @stellari for adding this problem and creating all test cases.

题目要求取两个链表的交点，而且时间复杂度必须是O(n)，所以就不能用嵌套循环的方法。用了如下方法，先计算两个链表的各自长度，将长链表节点向下移动两链表长度差，再计算。

代码

/** * 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 Aindex_length=getLength(headA);        int Bindex_length=getLength(headB);        int dis=Math.abs(Aindex_length-Bindex_length);                                            ListNode Aindex=headA;        ListNode Bindex=headB;        if(Aindex_length>=Bindex_length){          for(int i=0;i<dis;i++){              Aindex=Aindex.next;          }            while(Bindex!=null){              if(Aindex.val==Bindex.val){return Aindex;}              else{                  Aindex=Aindex.next;                  Bindex=Bindex.next;                 }          }        }                Aindex=headA;         Bindex=headB;      if(Aindex_length<Bindex_length){          for(int i=0;i<dis;i++){              Bindex=Bindex.next;          }            while(Aindex!=null){              if(Aindex.val==Bindex.val){return Aindex;}              else{                  Aindex=Aindex.next;                  Bindex=Bindex.next;                 }          }        }                        return null;    }    public int getLength(ListNode head){        ListNode index=head;        int length=1;        while(index.next!=null){            index=index.next;            length++;        }        return length;    }}

代码下载：https://github.com/jimenbian/GarvinLeetCode

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