2 Add Two Numbers 两数相加
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.
题目描述:
给出两个 非空 的链表用来表示两个非负的整数。其中，它们各自的位数是按照 逆序 的方式存储的，并且它们的每个节点只能存储 一位 数字。
如果，我们将这两个数相加起来，则会返回一个新的链表来表示它们的和。
您可以假设除了数字 0 之外，这两个数都不会以 0 开头。
示例 :
输入：(2 -> 4 -> 3) + (5 -> 6 -> 4)
输出：7 -> 0 -> 8
原因：342 + 465 = 807
思路:
类似加法, 设置一个进位记录是否有进位, 对链表逐位进行加减, 长度不足的用 0代替
时间复杂度O(n), 空间复杂度O(n)
代码:
C++:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution
{
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2)
{
ListNode* result = new ListNode(0);
ListNode* r = result;
int c = 0;
while (l1 or l2)
{
int x = l1 ? l1 -> val : 0, y = l2 ? l2 -> val : 0;
r -> next = new ListNode((c + x + y) % 10);
c = (c + x + y) / 10;
r = r -> next;
l1 = l1 ? l1 -> next : nullptr;
l2 = l2 ? l2 -> next : nullptr;
}
if (c > 0) r -> next = new ListNode(c);
return result -> next;
}
};
Java:
/**
* 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 result = new ListNode(0);
int c = 0;
ListNode r = result;
while (l1 != null || l2 != null) {
int x = l1 != null ? l1.val : 0, y = l2 != null ? l2.val : 0;
r.next = new ListNode((c + x + y) % 10);
c = (c + x + y) / 10;
r = r.next;
l1 = l1 != null ? l1.next : null;
l2 = l2 != null ? l2.next : null;
}
if (c > 0) r.next = new ListNode(c);
return result.next;
}
}
Python:
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
c, result = 0, ListNode(0)
r = result
while l1 or l2:
x, y = l1.val if l1 else 0, l2.val if l2 else 0
r.next = ListNode((c + x + y) % 10)
r = r.next
c = (c + x + y) // 10
l1, l2 = l1.next if l1 else None, l2.next if l2 else None
if c > 0:
r.next = ListNode(c)
return result.next
