add-two-numbers-ii
注意:有一种好的方法,是将链表倒转,然后依次相加。
但是,按照题目要求,用了不改变原链表的方法。
就是将两个链表增加到相同长度,然后递归相加,子函数返回后处理进位。
https://leetcode.com/problems/add-two-numbers-ii/
package com.company;
import java.util.*;
class ListNode {
int val;
ListNode next;
ListNode(int x) { val = x; }
}
class Solution {
ListNode addTwo(ListNode l1, ListNode l2) {
//System.out.printf("add two %d, %d \n", l1.val, l2.val);
ListNode ret = new ListNode(l1.val + l2.val);
if (l1.next != null && l2.next != null) {
ret.next = addTwo(l1.next, l2.next);
ret.val += ret.next.val / 10;
ret.next.val = ret.next.val % 10;
}
else {
ret.next = null;
}
return ret;
}
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
int l1len = 0;
ListNode ln = l1;
while (ln != null) {
l1len++;
ln = ln.next;
}
int l2len = 0;
ln = l2;
while (ln != null) {
l2len++;
ln = ln.next;
}
if (l1len < l2len) {
ln = l1;
l1 = l2;
l2 = ln;
l1len = l1len ^ l2len;
l2len = l1len ^ l2len;
l1len = l1len ^ l2len;
}
for (int i=0; i<l1len-l2len; i++) {
ln = new ListNode(0);
ln.next = l2;
l2 = ln;
}
ln = addTwo(l1, l2);
if (ln.val >= 10) {
ListNode newHead = new ListNode(ln.val / 10);
ln.val = ln.val % 10;
newHead.next = ln;
ln = newHead;
}
return ln;
}
}
public class Main {
public static void main(String[] args) {
System.out.println("Hello!");
Solution solution = new Solution();
ListNode l1 = new ListNode(7);
ListNode l11 = new ListNode(2);
ListNode l12 = new ListNode(4);
ListNode l13 = new ListNode(3);
l1.next = l11;
l11.next = l12;
l12.next = l13;
ListNode l2 = new ListNode(5);
ListNode l21 = new ListNode(6);
ListNode l22 = new ListNode(4);
l2.next = l21;
l21.next = l22;
ListNode ret = solution.addTwoNumbers(l1, l2);
System.out.printf("Get ret: \n");
while (ret != null) {
System.out.printf("%d", ret.val);
ret = ret.next;
}
System.out.println();
/*Iterator<List<Integer>> iterator = ret.iterator();
while (iterator.hasNext()) {
Iterator iter = iterator.next().iterator();
while (iter.hasNext()) {
System.out.printf("%d,", iter.next());
}
System.out.println();
}*/
System.out.println();
}
}
add-two-numbers-ii的更多相关文章
- [LeetCode] 445. Add Two Numbers II 两个数字相加之二
You are given two linked lists representing two non-negative numbers. The most significant digit com ...
- LeetCode 445. 两数相加 II(Add Two Numbers II)
445. 两数相加 II 445. Add Two Numbers II 题目描述 给定两个非空链表来代表两个非负整数.数字最高位位于链表开始位置.它们的每个节点只存储单个数字.将这两数相加会返回一个 ...
- 445. Add Two Numbers II - LeetCode
Question 445. Add Two Numbers II Solution 题目大意:两个列表相加 思路:构造两个栈,两个列表的数依次入栈,再出栈的时候计算其和作为返回链表的一个节点 Java ...
- LeetCode 445 Add Two Numbers II
445-Add Two Numbers II You are given two linked lists representing two non-negative numbers. The mos ...
- [LeetCode] Add Two Numbers II 两个数字相加之二
You are given two linked lists representing two non-negative numbers. The most significant digit com ...
- LeetCode Add Two Numbers II
原题链接在这里:https://leetcode.com/problems/add-two-numbers-ii/ 题目: You are given two linked lists represe ...
- 【LeetCode445】 Add Two Numbers II★★
题目描述: 解题思路: 给定两个链表(代表两个非负数),数字的各位以正序存储,将两个代表数字的链表想加获得一个新的链表(代表两数之和). 如(7->2->4->3)(7243) + ...
- 【LeetCode】445. Add Two Numbers II 解题报告(Python & C++)
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 先求和再构成列表 使用栈保存节点数字 类似题目 日期 ...
- 445. Add Two Numbers II ——while s1 or s2 or carry 题目再简单也要些测试用例
You are given two linked lists representing two non-negative numbers. The most significant digit com ...
- [Swift]LeetCode445. 两数相加 II | Add Two Numbers II
You are given two non-empty linked lists representing two non-negative integers. The most significan ...
随机推荐
- SQL Server DATEADD() 函数
SQL Server Date 函数 定义和用法 DATEADD() 函数在日期中添加或减去指定的时间间隔. 语法 DATEADD(datepart,number,date) date 参数是合法的日 ...
- hadoop 技巧
通过--config指定不同的集群 bin/hadoop --config ./conf_time/ dfs -ls /user/rd/*/for_*/ip_table/output/ rd下是都读写 ...
- 修改tomcat 启动45秒
当我们需要增加Tomcat的启动时间,修改方法如下:
- JQuery,UIbootstrap风格弹出层
<!DOCTYPE html><html xmlns="http://www.w3.org/1999/xhtml"><head> <met ...
- z-index兼容问题:关于ie6/7下的z-index
z-index这个属性其实在挺多地方都会用到,在百度上搜索也有大量关于z-index的篇幅去阐述这个属性,特别是在ie6下的z-index处理有更多的相关文章,本文就不再围绕z-index这一属性的基 ...
- 【POJ】【1704】Georgia and Bob
组合游戏 Nim游戏的一个变形 题解请看金海峰的博客 以下为引用: 分析:我们把棋子按位置升序排列后,从后往前把他们两两绑定成一对.如果总个数是奇数,就把最前面一个和边界(位置为0)绑定. 在同一对棋 ...
- Codeforces Round #363 (Div. 2)->A. Launch of Collider
A. Launch of Collider time limit per test 2 seconds memory limit per test 256 megabytes input standa ...
- epoll 知识总结
poll/select/epoll 对比 http://www.cnblogs.com/apprentice89/p/3234677.html ---有待继续学习 http://blog.chi ...
- 提升SQL Server速度整理索引碎片
转载:http://wenku.baidu.com/view/f64c8a707fd5360cba1adbea.html SQL Server2005索引碎片分析和解决方法 毫无疑问,给表添加索引 ...
- UVALive 6525
二分图最大匹配 #include<cstdio> #include<iostream> #include<cstring> #define MAX 10010 us ...