2. Add Two Numbers(链表尾插法)
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.
/**
* 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) {
ListNode list = null, s = null;
int sum = 0;
while (l1 != null || l2 != null) {
sum /= 10;
if (l1 != null) {
sum += l1.val;
l1 = l1.next;
}
if (l2 != null) {
sum += l2.val;
l2 = l2.next;
}
if (list == null) {
list = new ListNode(sum % 10);
s = list;
} else {
ListNode node = new ListNode(sum % 10);
s.next = node;
s = s.next;
}
}
if (sum / 10 == 1) {
s.next = new ListNode(1);
}
return list;
}
}
Debug code in playground:
/* -----------------------------------
* WARNING:
* -----------------------------------
* Your code may fail to compile
* because it contains public class
* declarations.
* To fix this, please remove the
* "public" keyword from your class
* declarations.
*/
/**
* 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) {
ListNode list = null, s = null;
int sum = 0;
while (l1 != null || l2 != null) {
sum /= 10;
if (l1 != null) {
sum += l1.val;
l1 = l1.next;
}
if (l2 != null) {
sum += l2.val;
l2 = l2.next;
}
if (list == null) {
list = new ListNode(sum % 10);
s = list;
} else {
ListNode node = new ListNode(sum % 10);
s.next = node;
s = s.next;
}
}
if (sum / 10 == 1) {
s.next = new ListNode(1);
}
return list;
}
}
public class MainClass {
public static int[] stringToIntegerArray(String input) {
input = input.trim();
input = input.substring(1, input.length() - 1);
if (input.length() == 0) {
return new int[0];
}
String[] parts = input.split(",");
int[] output = new int[parts.length];
for(int index = 0; index < parts.length; index++) {
String part = parts[index].trim();
output[index] = Integer.parseInt(part);
}
return output;
}
public static ListNode stringToListNode(String input) {
// Generate array from the input
int[] nodeValues = stringToIntegerArray(input);
// Now convert that list into linked list
ListNode dummyRoot = new ListNode(0);
ListNode ptr = dummyRoot;
for(int item : nodeValues) {
ptr.next = new ListNode(item);
ptr = ptr.next;
}
return dummyRoot.next;
}
public static String listNodeToString(ListNode node) {
if (node == null) {
return "[]";
}
String result = "";
while (node != null) {
result += Integer.toString(node.val) + ", ";
node = node.next;
}
return "[" + result.substring(0, result.length() - 2) + "]";
}
public static void main(String[] args) throws IOException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
String line;
while ((line = in.readLine()) != null) {
ListNode l1 = stringToListNode(line);
line = in.readLine();
ListNode l2 = stringToListNode(line);
ListNode ret = new Solution().addTwoNumbers(l1, l2);
String out = listNodeToString(ret);
System.out.print(out);
}
}
}
这个不要理所当然想成了头插法,看到测试代码才知道是尾插法,返回的ListNode也是需要尾插法的。
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CSDN博客地址:https://blog.csdn.net/qq_34115899