LeetCode 2. Add Two Numbers 原创Java参考解答
问题描述
https://leetcode.com/problems/add-two-numbers/
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
解题思路
题目是求做一个以两个链表l1, l2的每一位节点之和作为每一个节点值的新链表。两个链表从左往右，想象成个位、十位、百位相加……该进位时候需要进位。
破此题的关键点：由“个位”节点开始做加法构造新链表，注意分情况计算新链表每一个位置节点的值：两个链表都还有节点时 ((l1.val + l2.val + carry) % 10)、只剩其中一个链表有节点时 ((l1.val + carry) % 10 || (l2.val + carry) % 10)、两个链表都没有节点时 (carry)。
参考代码
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(0);
ListNode current = dummy;
int carry = 0;
while (l1 != null && l2 != null) {
int result = (l1.val + l2.val + carry) % 10;
ListNode newNode = new ListNode(result);
carry = (l1.val + l2.val + carry) / 10;
current.next = newNode;
current = newNode;
l1 = l1.next;
l2 = l2.next;
}
while (l1 != null) {
int result = (l1.val + carry) % 10;
ListNode newNode = new ListNode(result);
carry = (l1.val + carry) / 10;
current.next = newNode;
current = newNode;
l1 = l1.next;
}
while (l2 != null) {
int result = (l2.val + carry) % 10;
ListNode newNode = new ListNode(result);
carry = (l2.val + carry) / 10;
current.next = newNode;
current = newNode;
l2 = l2.next;
}
if (carry != 0) {
ListNode newNode = new ListNode(carry);
current.next = newNode;
current = newNode;
}
return dummy.next;
}
}