Leetcode2. Add Two Numbers
原创
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题目:
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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
思路:
没啥思路,就一位一位对应加就行,注意进位。
代码写得比较渣,但是好理解,也AC了。
import java.util.LinkedList;
class ListNode {
int val;
ListNode next;
ListNode(int x) { val = x; }
}
public class Test {
public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
int carry = 0;
LinkedList<Integer> list = new LinkedList();
while(l1 != null && l2 != null) {
int a = l1.val;
int b = l2.val;
int result = a + b + carry;
if(result >= 10) {
result %= 10;
carry = 1;
}else {
carry = 0;
}
list.add(result);
l1 = l1.next;
l2 = l2.next;
}
while(l1 != null) {
int result = l1.val + carry;
if(result >= 10) {
result %= 10;
carry = 1;
}else {
carry = 0;
}
list.add(result);
l1 = l1.next;
}
while(l2 != null) {
int result = l2.val + carry;
if(result >= 10) {
result %= 10;
carry = 1;
}else {
carry = 0;
}
list.add(result);
l2 = l2.next;
}
if(carry != 0) {
list.add(carry);
}
// 创造返回链表
ListNode head = new ListNode(list.removeFirst());
ListNode result = head;
while(!list.isEmpty()) {
head.next = new ListNode(list.removeFirst());
head = head.next;
}
return result;
}
public static ListNode createTestCase(int[] arr) {
ListNode head = new ListNode(-1);
ListNode h = head;
for(int i = 0; i < arr.length; i++) {
head.next = new ListNode(arr[i]);
head = head.next;
}
return h.next;
}
public static void print(ListNode head) {
while(head != null) {
System.out.print(head.val + " ");
head = head.next;
}
System.out.println();
}
public static void main(String[] args) {
// Test case
int[] arr1 = {2,4,5,1};
int[] arr2 = {5,6,4};
ListNode l1 = createTestCase(arr1);
ListNode l2 = createTestCase(arr2);
ListNode l3 = addTwoNumbers(l1, l2);
print(l3);
}
}