https://leetcode.com/problems/add-two-numbers/description/
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.
題目是給定兩個非空倒敘排列的鏈表,要求將兩個鏈表相加形成一個新的鏈表。新的鏈表同樣是倒敘排列。如2-4-3 + 5-6-4, 即爲342+465 = 708 , 所得鏈表需要爲8-0-7.
本題不用考慮倒序問題,因爲正着加和反着加結果是一樣的,結果所得鏈表也爲倒序。
/**
* 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) {
int sum = 0;
ListNode temp = new ListNode(0);//初始化新鏈表,用於操作
ListNode temp1 = temp;//保存新鏈表頭部,用於返回。
while (l1 != null || l2 != null) {
sum /= 10;//處理進位問題
if (null != l1) {//相加
sum += l1.val;
l1 = l1.next;
}
if (null != l2) {
sum += l2.val;
l2 = l2.next;
}
temp.next = new ListNode(sum % 10);//保存相加結果的個位數
temp = temp.next;
}
if(sum/10 == 1)//如果此時一個鏈表已經爲空,即兩個鏈表長度不同,則需要處理最後一位的問題。(如倒數第二位進位爲9,最後一個鏈表末尾數爲9,則9+9=18,進位1,放到最後一位)
temp.next = new ListNode(1);
return temp1.next;//從temp的next開始返回
}
}
放一個相似解法,和上面區別不大public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode head = new ListNode(0);
ListNode result = head;
int carry = 0;
while (l1 != null || l2 != null || carry > 0) {
int resVal = (l1 != null? l1.val : 0) + (l2 != null? l2.val : 0) + carry;
result.next = new ListNode(resVal % 10);
carry = resVal / 10;
l1 = (l1 == null ? l1 : l1.next);
l2 = (l2 == null ? l2 : l2.next);
result = result.next;
}
return head.next;
}