【LeetCode】445. Add Two Numbers II 解题报告（Python & C++）

作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

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题目地址：https://leetcode.com/problems/add-two-numbers-ii/description/

题目描述

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first 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.

Follow up:

What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:

Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 8 -> 0 -> 7

题目大意

有两个链表，分别是正序的十进制数字。现在要求两个十位数字的和，要求返回的结果也是链表。

解题方法

先求和再构成列表

选择easy模式的话，可以直接先求出和，再构建链表。

这么做的前提是python中没有最大的整数，所以无论怎么着不会越界！

Python代码如下：

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        num1 = ''
        num2 = ''
        while l1:
            num1 += str(l1.val)
            l1 = l1.next
        while l2:
            num2 += str(l2.val)
            l2 = l2.next
        add = str(int(num1) + int(num2))
        head = ListNode(add[0])
        answer = head
        for i in range(1, len(add)):
            node = ListNode(add[i])
            head.next = node
            head = head.next
        return answer

使用栈保存节点数字

可以采用翻转后变成 Add Two Numbers 的题目，但是题目说了最好别那么做。那么可以使用一个栈来完成类似的操作。上一个题目的结果也是数字的倒序的，而这个题要求结果是正序，那么加的过程中采用头插法，一直向头部插入新的节点就好了。

步骤非常类似Add Two Numbers。

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        stack1 = []
        stack2 = []
        while l1:
            stack1.append(l1.val)
            l1 = l1.next
        while l2:
            stack2.append(l2.val)
            l2 = l2.next
        answer = None
        carry = 0
        while stack1 and stack2:
            add = stack1.pop() + stack2.pop() + carry
            carry = 1 if add >= 10 else 0
            temp = answer
            answer = ListNode(add % 10)
            answer.next = temp
        l = stack1 if stack1 else stack2
        while l:
            add = l.pop() + carry
            carry = 1 if add >= 10 else 0
            temp = answer
            answer = ListNode(add % 10)
            answer.next = temp
        if carry:
            temp = answer
            answer = ListNode(1)
            answer.next = temp
        return answer

二刷的时候使用的C++的vector来存储的每个节点的值，然后再做的加法。每次构建新节点使用的头插法，所以结果同样是题目要求的倒序。

C++代码如下：

/**
 * 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) {
        vector<int> v1, v2;
        while (l1) {
            v1.push_back(l1->val);
            l1 = l1->next;
        }
        while (l2) {
            v2.push_back(l2->val);
            l2 = l2->next;
        }
        const int M = v1.size(), N = v2.size();
        int i1 = M - 1, i2 = N - 1;
        int carry = 0;
        ListNode* dummy = new ListNode(0);
        ListNode* cur = dummy;
        while (i1 >= 0 || i2 >= 0 || carry) {
            int add = (i1 >= 0 ? v1[i1] : 0) + (i2 >= 0 ? v2[i2] : 0) + carry;
            carry = add >= 10 ? 1 : 0;
            add %= 10;
            ListNode* cur = new ListNode(add);
            cur->next = dummy->next;
            dummy->next = cur;
            if (i1 >= 0)
                --i1;
            if (i2 >= 0)
                --i2;
        }
        return dummy->next;
    }
};

类似题目

2. Add Two Numbers
989. Add to Array-Form of Integer

日期

2018 年 2 月 26 日
2019 年 2 月 22 日 —— 这周结束了