題目

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.

題目要求取兩個(gè)鏈表的交點(diǎn)，而且時(shí)間復(fù)雜度必須是O(n)，所以就不能用嵌套循環(huán)的方法。用了如下方法，先計(jì)算兩個(gè)鏈表的各自長(zhǎng)度，將長(zhǎng)鏈表節(jié)點(diǎn)向下移動(dòng)兩鏈表長(zhǎng)度差，再計(jì)算。

代碼

/*** 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;} }



代碼下載：https://github.com/jimenbian/GarvinLeetCode

/********************************

* 本文來(lái)自博客 ?“李博Garvin“

* 轉(zhuǎn)載請(qǐng)標(biāo)明出處:http://blog.csdn.net/buptgshengod

******************************************/

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