原题：
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.
Example
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8 Explanation: 342 + 465 = 807.
给出的类和方法参数
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
}
}
大致意思就是输入两个链表，每个链表里node的值分别是两个数字的倒序排列，最终输出同样用链表输出两个数字和的倒序排列
先完成一个版本
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
int num1, num2, num0;
ListNode l = null;
if (l1 != null && l2 != null &&
l1.next != null && l2.next != null &&
l1.next.next != null && l2.next.next != null) {
num1 = l1.val + l1.next.val * 10 + l1.next.next.val * 100;
num2 = l2.val + l2.next.val * 10 + l2.next.next.val * 100;
num0 = num1 + num2;
l = new ListNode(num0 % 10);
l.next = new ListNode((num0 / 10) % 10);
l.next.next = new ListNode(num0 / 100);
}
return l;
}
}
显然中间的if以及链表的遍历太恶心了，而且根本就不满足题目要求，因为根本没说是百位数，而且位数可以不一样，因此对于应该采用正常链表的遍历方法，分别获取数字，求和之后倒序写入链表返回，重新写下
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
int num1 = 0, num2 = 0, num0;
int i = 0;
while (l1 != null) {
num1 += l1.val * Math.pow(10, i);
i++;
l1 = l1.next;
}
int j = 0;
while (l2 != null) {
num2 += l2.val * Math.pow(10, j);
j++;
l2 = l2.next;
}
num0 = num1 + num2;
ListNode l = null;
l = new ListNode(num0 % 10);
l.next = new ListNode((num0 / 10) % 10);
l.next.next = new ListNode(num0 / 100);
return l;
}
}
提交之后，报了一个错误
9 / 1562 test cases passed. Status: Wrong Answer Submitted: 0 minutes ago Input: [0] [0] Output: [0,0,0] Expected: [0]
这里不满足的case为输入两个0，那就加一个判断，如果和为0，就返回0
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
int num1 = 0, num2 = 0, num0;
int i = 0;
while (l1 != null) {
num1 += l1.val * Math.pow(10, i);
i++;
l1 = l1.next;
}
int j = 0;
while (l2 != null) {
num2 += l2.val * Math.pow(10, j);
j++;
l2 = l2.next;
}
num0 = num1 + num2;
ListNode l = null;
if (num0 == 0) {
l = new ListNode(0);
} else {
l = new ListNode(num0 % 10);
l.next = new ListNode((num0 / 10) % 10);
l.next.next = new ListNode(num0 / 100);
}
return l;
}
}
再次提交，还是错误
10 / 1562 test cases passed. Status: Wrong Answer Submitted: 0 minutes ago Input: [0] [1] Output: [1,0,0] Expected: [1]
原来刚只将输入的数字的遍历方式改了，输出l对象还是当成三位数了，修改为链表添加值
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
int num1 = 0, num2 = 0, num0;
int i = 0;
while (l1 != null) {
num1 += l1.val * Math.pow(10, i);
i++;
l1 = l1.next;
}
int j = 0;
while (l2 != null) {
num2 += l2.val * Math.pow(10, j);
j++;
l2 = l2.next;
}
num0 = num1 + num2;
ListNode l = new ListNode(0);
ListNode head = l;
while (num0 != 0) {
l.val = num0 % 10;
num0 /= 10;
if (num0 != 0) {
l.next = new ListNode(0);
l = l.next;
} else {
l.next = null;
}
}
return head;
}
}
还是错误
787 / 1562 test cases passed. Status: Wrong Answer Submitted: 0 minutes ago Input: [9] [1,9,9,9,9,9,9,9,9,9] Output: [0,-4,-6,-3,-8,-4,-7,-4,-1,-2] Expected: [0,0,0,0,0,0,0,0,0,0,1]
仔细看看，输出的结果显示是越界了，原因是程序输入的是链表，存储的是倒序的数字，我这里的处理方法是，先把每个数字正序过来，求和，然后再倒序输出，这样就会出现一种错误场景，正序的数字有可能越界，而倒序的数字没有越界，这样不影响程序输入和输出，因为输入输出是链表，因此问题还在于程序实现的方法不行
本来是不想计较两个数字位数不一致，所以直接计算和，倒序，如此看来还是得每一位求和，然后写入链表
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode l = new ListNode(0);
ListNode head = l;
int sum = 0, tmp = 0;
//l1 : 9 9
//l2 : 1 9 9 9
//l.next : 0 9 0 0 1
while (l1 != null && l2 != null) {
sum = l1.val + l2.val + tmp;
l.next = new ListNode(sum % 10);
tmp = sum / 10;
l = l.next;
l1 = l1.next;
l2 = l2.next;
}
while (l2 != null) {
sum = l2.val + tmp;
l.next = new ListNode(sum % 10);
tmp = sum / 10;
l = l.next;
l2 = l2.next;
}
while (l1 != null) {
sum =l1.val + tmp;
l.next = new ListNode(sum % 10);
tmp = sum / 10;
l = l.next;
l1 = l1.next;
}
if (tmp == 1) {
l.next = new ListNode(tmp);
}
return head.next;
}
}
提交验证无误
Submission Detail 1562 / 1562 test cases passed. Status: Accepted Runtime: 69 ms Submitted: 0 minutes ago
每一对相同位置的数字求和，从l.next开始写，sum % 10为当前位置做和的值，sum / 10决定是否要前一位+1
当位数少的那个数遍历完了，就剩下位数多的那个数，照葫芦画票，摩10的值写入链表，除10的商0或者1都加到高位（倒序的next）
当位数多的这个数遍历完了，while循环就退出了，但是sum是否要进位还不知道，因此最后还要判断一下，如果满10了，高位的1还要写入新的链表值里
最后因为是从l.next开始写的，因此遍历head也要从next开始
有人给出了递归的写法
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
if (l1 == null || l2 == null) {
return l1 == null ? l2 : l1;
}
int value = l1.val + l2.val;
ListNode result = new ListNode(value % 10);
result.next = addTwoNumbers(l1.next, l2.next);
if (value >= 10) {
result.next = addTwoNumbers(new ListNode(value / 10), result.next);
}
return result;
}
}
