LeetCode 2. Add Two Numbers [Medium]

链表加法。

题目来源:https://leetcode.com/problems/add-two-numbers

中文版:https://leetcode-cn.com/problems/add-two-numbers/

题目难度:Medium

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 contains a single digit. Add the two numbers and return the sum as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

解答1[Java]:

核心思想

代码

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class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode head = null, tail = null;
int carry = 0;
while (l1 != null || l2 != null) {
int n1 = l1 != null ? l1.val : 0;
int n2 = l2 != null ? l2.val : 0;
int sum = n1 + n2 + carry;
if (head == null) {
head = tail = new ListNode(sum % 10);
} else {
tail.next = new ListNode(sum % 10);
tail = tail.next;
}
carry = sum / 10;
if (l1 != null) {
l1 = l1.next;
}
if (l2 != null) {
l2 = l2.next;
}
}
if (carry > 0) {
tail.next = new ListNode(carry);
}
return head;
}
}

解答2[Java]:递归思路

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public class Problem2Solution2 {
class ListNode {
int val;
ListNode next;

ListNode() {
}

ListNode(int val) {
this.val = val;
}

ListNode(int val, ListNode next) {
this.val = val;
this.next = next;
}
}

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
if (l1 == null && l2 == null) {
return null;
}

if (l1 == null) {
l1 = new ListNode(0);
}

if (l2 == null) {
l2 = new ListNode(0);
}

int rval = l1.val + l2.val;
if (rval >= 10) {
rval = rval % 10;
if (l1.next != null) {
l1.next.val += 1;
} else {
l1.next = new ListNode(1);
}
}

return new ListNode(rval, addTwoNumbers(l1.next, l2.next));
}
}

进阶:链表中的数是正向存储的

思路1:给较短链表的长度和较长链表的长度补齐,补的高位节点的值都设为0。