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
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:
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