英文:
String to a singly linked list
问题
我正在尝试将一个字符串转换为链表，其中每个数字都在单独的节点中。
我着手尝试调试它，但似乎找不到我的逻辑有什么问题。
我一直得到一个奇怪的两位数，而不一定是字符串中出现的数字。
请注意，ListNode 是我创建新节点对象的类。
String number = "807";int size = number.length();int pos = 0;ListNode dummyhead = new ListNode();ListNode curr = dummyhead;while (size > 0){curr.next = new ListNode(number.charAt(pos));pos++;size--;curr = curr.next;}return dummyhead.next;
英文:
I am trying to convert a string to a linked list where each digit is in a separate node.
I set on it trying to debug it but I can't seem to find what's wrong with my logic.
I keep getting a weird 2-digit number in each node and not necessarily even the digits that appear in the string.
Please note the ListNode is the class I create a new node object with.
String number = "807";int size = number.length();int pos = 0;ListNode dummyhead = new ListNode();ListNode curr = dummyhead;while (size > 0){curr.next = new ListNode(number.charAt(pos));pos++;size--;curr = curr.next;}return dummyhead.next;
答案1
得分: 2
I think you're on the right track. The method works fine, but it seems you are not iterating the list right. Here is how I tested your code:
public class ListNode {ListNode next;char data;public ListNode(char data) {this.data = data;}public ListNode() {}}
private static ListNode getList(String number) {int size = number.length();int pos = 0;ListNode dummyhead = new ListNode();ListNode curr = dummyhead;while (size > 0) {curr.next = new ListNode(number.charAt(pos));pos++;size--;curr = curr.next;}return dummyhead.next;}private static String printList(ListNode head) {ListNode n = head;StringBuilder sb = new StringBuilder();while (n != null) {sb.append(n.data + "-");n = n.next;}return sb.toString();}public static void main(String[] args) {String number = "807";System.out.println(printList(getList(number)));}
Output:
8-0-7-
英文:
I think you're on the right track. The method works fine, but it seems you are not iterating the list right. Here is how I tested your code:
public class ListNode{ListNode next;char data;public ListNode(char data) {this.data = data;}public ListNode() {}}
private static ListNode getList(String number){int size = number.length();int pos = 0;ListNode dummyhead = new ListNode();ListNode curr = dummyhead;while (size > 0){curr.next = new ListNode(number.charAt(pos));pos++;size--;curr = curr.next;}return dummyhead.next;}private static String printList(ListNode head) {ListNode n = head;StringBuilder sb = new StringBuilder();while(n != null) {sb.append(n.data+"-");n = n.next;}return sb.toString();}public static void main(String[] args) {String number = "807";System.out.println(printList(getList(number)));}
Output:
8-0-7-
答案2
得分: 1
"charAt" 返回指定位置的字符。char 是一种数字类型，但它表示它所代表的字符的ASCII地址。
"807".charAt(0)
返回 56，因为这是 8 的ASCII值。
我怀疑您在 ListNode 内部有一个保存 char 到常规 int 的 int data 字段。
因此，您的 "807" 将被转换为数字 56、48、55 的列表。
显然，您希望将 ""8"" 保存到节点中，所以可以使用以下方式：
while (size > 0){//对于 pos=0，这将字符串 ""8"" 转换为整数 8：Integer n = Integer.valueOf(number.substring(pos, pos+1));curr.next = new ListNode(n);
或者，如Majed在他的答案中建议的那样，将 ListNode 内部的 data 字段类型更改为 char。
英文:
charAt returns the character of the specified position. A char is a numeric type, but it represents the ascii address of the character it represents.
"807".charAt(0)
returns 56 because that is the ascii value of 8.
I suspect that you have a int data field inside ListNode, which saves that char into a regular int.
Your "807" will thus be transformed into a list of the numbers 56, 48, 55.
You apparently want to save that "8" into the node, so use
while (size > 0){//for pos=0, this converts the string "8" to the integer 8:Integer n = Integer.valueOf(number.substring(pos, pos+1));curr.next = new ListNode(n);
Or, as Majed suggests in his answer, change the type of your data field inside of ListNode to char.
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论