我正在尝试添加两个非负数，其中的数字以相反的顺序存储在两个单独的链接列表中。答案也应该是一个链表，数字反转，没有尾随零。
据我所知，有一种方法可以通过添加数字和每次维持一个进位来解决这个问题，但我试图通过对数字使用加法运算来解决它。
这是我的代码：
/**
* Definition for singly-linked list.
* class ListNode {
* public int val;
* public ListNode next;
* ListNode(int x) { val = x; next = null; }
* }
*/
public class Solution {
public ListNode addTwoNumbers(ListNode a, ListNode b) {
if(a==null || b==null){
return null;
}
String num1 = "";
String num2 = "";
ListNode temp1 = a;
ListNode temp2 = b;
while(temp1!=null){
num1 = num1+Integer.toString(temp1.val);
temp1 = temp1.next;
}
new StringBuilder(num1).reverse().toString();
double value1 = Double.parseDouble(num1);
while(temp2!=null){
num2 = num2+Integer.toString(temp2.val);
temp2 = temp2.next;
}
new StringBuilder(num2).reverse().toString();
double value2 = Double.parseDouble(num2);
double result = value1+value2;
String res = String.format("%.0f",result);
ListNode first_node = new ListNode(Character.getNumericValue(res.charAt(0)));
ListNode ans = first_node;
for(int j=1;j<res.length();j++){
ListNode node = new ListNode(Character.getNumericValue(res.charAt(j)));
add(node,ans);
}
return ans;
}
public void add(ListNode node, ListNode ans){
ListNode temp;
temp = ans;
ans = node;
ans.next = temp;
}
}
我的代码一直给出错误的答案。任何人都可以指出错误吗？
答案 0 :(得分：0)
您的方法不正确且间接。 你正在尝试使用浮点运算来做大数字算术。 结果 - 计算错误。
我们有：
List<Integer> firstNumber;
List<Integer> secondNumber;
让假设 firstNumber > secondNumber。
试试这个算法：
List<Integer> result = new ArrayList<>();
int i = 0;
int appendix = 0;
for (; i < secondNumber.size(); i++) {
int sum = firstNumber.get(i) + secondNumber.get(i) + appendix;
result.append(sum % 10);
appendix = sum / 10;
}
for (; i < firstNumber.size(); i++) {
int sum = firstNumber.get(i) + appendix;
result.append(sum % 10);
appendix = sum / 10;
}
if (appendix != 0)
result.append(appendix);
return result;
答案 1 :(得分：0)
您的添加功能看起来不正确。您将以与预期相反的顺序获得编号。
此外，您的解决方案缺少问题的要点。如果你有一个有很多数字的数字你的方法将会失败（即使是双倍的限制，我认为它是2 ^ 1024）。链表表示允许数字更大。
正确的解决方案是在创建解决方案列表的同时，使用进位数同时迭代两个列表。如果这是作业或编码竞赛中的问题，那么您的解决方案将被判定为错误。
答案 2 :(得分：0)
您的add方法错误，它没有正确构建列表。以下是您的方法的最后部分应该如何看，没有add方法：
for(int j=1;j<res.length();j++){
ans.next = new ListNode(Character.getNumericValue(res.charAt(j)));;
ans = ans.next;
}
return first_node;
答案 3 :(得分：0)
在您的方法中，变量ans不会更新。你可以试试这个：
ans = add(node,ans);
并在您的add方法中，更改方法以返回ListNode ans
答案 4 :(得分：0)
您的方法并不简单，也不会给您预期的结果。
这是一个简单的方法，不需要太多解释，因为加法是整数的简单整数。
请注意，当两个整数之和大于9时，我继续前进，否则继续使用列表中下一个整数的总和。
class Node {
private Object data;
private Node next;
public Object getData() { return data; }
public void setData(Object data) { this.data = data; }
public Node getNext() { return next; }
public void setNext(Node next) { this.next = next; }
public Node(final Object data, final Node next) {
this.data = data;
this.next = next;
}
@Override
public String toString() { return "Node:[Data=" + data + "]"; }
}
class SinglyLinkedList {
Node start;
public SinglyLinkedList() { start = null; }
public void addFront(final Object data) {
// create a reference to the start node with new data
Node node = new Node(data, start);
// assign our start to a new node
start = node;
}
public void addRear(final Object data) {
Node node = new Node(data, null);
Node current = start;
if (current != null) {
while (current.getNext() != null) {
current = current.getNext();
}
current.setNext(node);
} else {
addFront(data);
}
}
public void deleteNode(final Object data) {
Node previous = start;
if (previous == null) {
return;
}
Node current = previous.getNext();
if (previous != null && previous.getData().equals(data)) {
start = previous.getNext();
previous = current;
current = previous.getNext();
return;
}
while (current != null) {
if (current.getData().equals(data)) {
previous.setNext(current.getNext());
current = previous.getNext();
} else {
previous = previous.getNext();
current = previous.getNext();
}
}
}
public Object getFront() {
if (start != null) {
return start.getData();
} else {
return null;
}
}
public void print() {
Node current = start;
if (current == null) {
System.out.println("SingleLinkedList is Empty");
}
while (current != null) {
System.out.print(current);
current = current.getNext();
if (current != null) {
System.out.print(", ");
}
}
}
public int size() {
int size = 0;
Node current = start;
while (current != null) {
current = current.getNext();
size++;
}
return size;
}
public Node getStart() {
return this.start;
}
public Node getRear() {
Node current = start;
Node previous = current;
while (current != null) {
previous = current;
current = current.getNext();
}
return previous;
}
}
public class AddNumbersInSinglyLinkedList {
public static void main(String[] args) {
SinglyLinkedList listOne = new SinglyLinkedList();
SinglyLinkedList listTwo = new SinglyLinkedList();
listOne.addFront(5);
listOne.addFront(1);
listOne.addFront(3);
listOne.print();
System.out.println();
listTwo.addFront(2);
listTwo.addFront(9);
listTwo.addFront(5);
listTwo.print();
SinglyLinkedList listThree = add(listOne, listTwo);
System.out.println();
listThree.print();
}
private static SinglyLinkedList add(SinglyLinkedList listOne, SinglyLinkedList listTwo) {
SinglyLinkedList result = new SinglyLinkedList();
Node startOne = listOne.getStart();
Node startTwo = listTwo.getStart();
int carry = 0;
while (startOne != null || startTwo != null) {
int one = 0;
int two = 0;
if (startOne != null) {
one = (Integer) startOne.getData();
startOne = startOne.getNext();
}
if (startTwo != null) {
two = (Integer) startTwo.getData();
startTwo = startTwo.getNext();
}
int sum = carry + one + two;
carry = 0;
if (sum > 9) {
carry = sum / 10;
result.addRear(sum % 10);
} else {
result.addRear(sum);
}
}
return result;
}
}
示例运行
Node:[Data=3], Node:[Data=1], Node:[Data=5]
Node:[Data=5], Node:[Data=9], Node:[Data=2]
Node:[Data=8], Node:[Data=0], Node:[Data=8]