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：$O(M + N)$
最后更新于
