Add Two Numbers

2.Add Two Numbers

from LeetCode

Description

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

Solution

My Accepted Solution Run Time(1562 test cases) A 70ms B 50ms

代码实现

A：在计算和之前就new出空节点

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        ListNode presult = new ListNode(0);        ListNode result = presult;        int carry = 0;        ListNode n1,n2;        for(n1 = l1,n2 = l2;n1 != null && n2!=null ;n1=n1.next)        {            int temp =(n1.val+n2.val+carry);             presult.val = temp%10;            carry = temp/10;            n2 = n2.next;            presult.next = new ListNode(0);            presult = presult.next;        }        while(n1 != null)        {            int temp = (n1.val+carry);            presult.val=temp%10;            carry = temp/10;            presult.next = new ListNode(0);            presult = presult.next;            n1 = n1.next;        }        while(n2 != null)        {            int temp = (n2.val+carry);            presult.val=temp%10;            carry = temp/10;            presult.next = new ListNode(0);            presult = presult.next;            n2 = n2.next;        }         if(n1 == null && n2 == null)             if(carry != 0)                {                    presult.val = carry;                    presult.next = new ListNode(0);                    presult = presult.next;                }        //删除多new出的节点        ListNode prev;        presult = result;        while(presult != null)        {            prev = presult;            presult=presult.next;            if(presult.val==0 &&presult.next == null )            {                    prev.next=null;                    break;            }        }            return result;    }}

B：在计算和之后new节点，并用结果调用构造函数

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        ListNode presult = new ListNode(0),result = presult;        int carry = 0;        ListNode n1,n2;        for(n1 = l1,n2 = l2;n1 != null && n2!=null ;n1=n1.next)        {            int temp =(n1.val+n2.val+carry);             presult.next = new ListNode(temp%10);             carry = temp/10;            n2 = n2.next;            presult = presult.next;        }        while(n1 != null)        {            int temp = (n1.val+carry);            presult.next = new ListNode(temp%10);            carry = temp/10;            presult = presult.next;            n1 = n1.next;        }        while(n2 != null)        {            int temp = (n2.val+carry);            presult.next = new ListNode(temp%10);            carry = temp/10;            presult = presult.next;            n2 = n2.next;        }        if(carry != 0)            presult.next = new ListNode(carry);        //第一个节点是多余的，是为了不破坏代码逻辑new出来的头节点        return result.next;    }}

日志：

我的算法学得真的不是很好，大一的时候就有同学参加各种集训队了，当我意思到自己比别人差的时候已经是现在了，大二快大三了。刷题就算是维持自己写代码的常态吧，然后把自己的答案记录在博客上观察自己的成长，并不断更新更优的解决方案。Add Two Numbers此题只是用到了ListNode，链表的数据结构。没有用到高深的算法。