【LeetCode】445. Add Two Numbers II 解題報告（Python & C++）

時間 2021-08-14 標籤 node python git app ide spa code orm leetcode

做者：負雪明燭
id： fuxuemingzhu
我的博客： http://fuxuemingzhu.cn/node

題目描述

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.git

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

Follow up:ide

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

Example:

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

題目大意

有兩個鏈表，分別是正序的十進制數字。如今要求兩個十位數字的和，要求返回的結果也是鏈表。code

解題方法

先求和再構成列表

選擇easy模式的話，能夠直接先求出和，再構建鏈表。orm

這麼作的前提是python中沒有最大的整數，因此不管怎麼着不會越界！ip

Python代碼以下：leetcode

# 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

使用棧保存節點數字

能夠採用翻轉後變成 Add Two Numbers 的題目，可是題目說了最好別那麼作。那麼能夠使用一個棧來完成相似的操做。上一個題目的結果也是數字的倒序的，而這個題要求結果是正序，那麼加的過程當中採用頭插法，一直向頭部插入新的節點就行了。

步驟很是相似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

二刷的時候使用的C++的vector來存儲的每一個節點的值，而後再作的加法。每次構建新節點使用的頭插法，因此結果一樣是題目要求的倒序。

C++代碼以下：

/** * 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;
    }
};

相似題目

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 —— 這周結束了