0445. 两数相加 II

题目地址(445. 两数相加 II)

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

题目描述

给你两个 非空 链表来代表两个非负整数。数字最高位位于链表开始位置。它们的每个节点只存储一位数字。将这两数相加会返回一个新的链表。

你可以假设除了数字 0 之外，这两个数字都不会以零开头。

 

进阶：

如果输入链表不能修改该如何处理？换句话说，你不能对列表中的节点进行翻转。

 

示例：

输入：(7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
输出：7 -> 8 -> 0 -> 7

前置知识

  • 链表

  • 栈

公司

  • 腾讯

  • 百度

  • 字节

思路

由于需要从低位开始加，然后进位。 因此可以采用栈来简化操作。 依次将两个链表的值分别入栈 stack1 和 stack2，然后相加入栈 stack，进位操作用一个变量 carried 记录即可。

最后根据 stack 生成最终的链表即可。

也可以先将两个链表逆置，然后相加，最后将结果再次逆置。

关键点解析

  • 栈的基本操作

  • carried 变量记录进位

  • 循环的终止条件设置成stack.length > 0 可以简化操作

  • 注意特殊情况， 比如 1 + 99 = 100

代码

  • 语言支持：JS，C++, Python3

JavaScript Code:

/*
 * @lc app=leetcode id=445 lang=javascript
 *
 * [445] Add Two Numbers II
 */
/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var addTwoNumbers = function (l1, l2) {
  const stack1 = [];
  const stack2 = [];
  const stack = [];

  let cur1 = l1;
  let cur2 = l2;
  let curried = 0;

  while (cur1) {
    stack1.push(cur1.val);
    cur1 = cur1.next;
  }

  while (cur2) {
    stack2.push(cur2.val);
    cur2 = cur2.next;
  }

  let a = null;
  let b = null;

  while (stack1.length > 0 || stack2.length > 0) {
    a = Number(stack1.pop()) || 0;
    b = Number(stack2.pop()) || 0;

    stack.push((a + b + curried) % 10);

    if (a + b + curried >= 10) {
      curried = 1;
    } else {
      curried = 0;
    }
  }

  if (curried === 1) {
    stack.push(1);
  }

  const dummy = {};

  let current = dummy;

  while (stack.length > 0) {
    current.next = {
      val: stack.pop(),
      next: null,
    };

    current = current.next;
  }

  return dummy.next;
};

C++ Code：

/**
 * 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) {
        auto carry = 0;
        auto ret = (ListNode*)nullptr;
        auto s1 = vector<int>();
        toStack(l1, s1);
        auto s2 = vector<int>();
        toStack(l2, s2);
        while (!s1.empty() || !s2.empty() || carry != 0) {
            auto v1 = 0;
            auto v2 = 0;
            if (!s1.empty()) {
                v1 = s1.back();
                s1.pop_back();
            }
            if (!s2.empty()) {
                v2 = s2.back();
                s2.pop_back();
            }
            auto v = v1 + v2 + carry;
            carry = v / 10;
            auto tmp = new ListNode(v % 10);
            tmp->next = ret;
            ret = tmp;
        }
        return ret;
    }
private:
    // 此处若返回而非传入vector，跑完所有测试用例多花8ms
    void toStack(const ListNode* l, vector<int>& ret) {
        while (l != nullptr) {
            ret.push_back(l->val);
            l = l->next;
        }
    }
};

// 逆置，相加，再逆置。跑完所有测试用例比第一种解法少花4ms
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        auto rl1 = reverseList(l1);
        auto rl2 = reverseList(l2);
        auto ret = add(rl1, rl2);
        return reverseList(ret);
    }
private:
    ListNode* reverseList(ListNode* head) {
        ListNode* prev = NULL;
        ListNode* cur = head;
        ListNode* next = NULL;
        while (cur != NULL) {
            next = cur->next;
            cur->next = prev;
            prev = cur;
            cur = next;
        }
        return prev;
    }

    ListNode* add(ListNode* l1, ListNode* l2) {
        ListNode* ret = nullptr;
        ListNode* cur = nullptr;
        int carry = 0;
        while (l1 != nullptr || l2 != nullptr || carry != 0) {
            carry += (l1 == nullptr ? 0 : l1->val) + (l2 == nullptr ? 0 : l2->val);
            auto temp = new ListNode(carry % 10);
            carry /= 10;
            if (ret == nullptr) {
                ret = temp;
                cur = ret;
            }
            else {
                cur->next = temp;
                cur = cur->next;
            }
            l1 = l1 == nullptr ? nullptr : l1->next;
            l2 = l2 == nullptr ? nullptr : l2->next;
        }
        return ret;
    }
};

Python Code:

# 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:
        def listToStack(l: ListNode) -> list:
            stack, c = [], l
            while c:
                stack.append(c.val)
                c = c.next
            return stack

        # transfer l1 and l2 into stacks
        stack1, stack2 = listToStack(l1), listToStack(l2)

        # add stack1 and stack2
        diff = abs(len(stack1) - len(stack2))
        stack1 = ([0]*diff + stack1 if len(stack1) < len(stack2) else stack1)
        stack2 = ([0]*diff + stack2 if len(stack2) < len(stack1) else stack2)
        stack3 = [x + y for x, y in zip(stack1, stack2)]

        # calculate carry for each item in stack3 and add one to the item before it
        carry = 0
        for i, val in enumerate(stack3[::-1]):
            index = len(stack3) - i - 1
            carry, stack3[index] = divmod(val + carry, 10)
            if carry and index == 0:
                stack3 = [1] + stack3
            elif carry:
                stack3[index - 1] += 1

        # transfer stack3 to a linkedList
        result = ListNode(0)
        c = result
        for item in stack3:
            c.next = ListNode(item)
            c = c.next

        return result.next

复杂度分析

其中 M 和 N 分别为两个链表的长度。

  • 时间复杂度：$O(M + N)$

  • 空间复杂度：$O(M + N)$

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