0445. 兩數相加 II · ttttt

# 0445. 兩數相加 II ## 題目地址(445. 兩數相加 II) <https://leetcode-cn.com/problems/add-two-numbers-ii/> ## 題目描述 ``` <pre class="calibre18">``` 給你兩個非空鏈表來代表兩個非負整數。數字最高位位于鏈表開始位置。它們的每個節點只存儲一位數字。將這兩數相加會返回一個新的鏈表。你可以假設除了數字 0 之外，這兩個數字都不會以零開頭。進階：如果輸入鏈表不能修改該如何處理？換句話說，你不能對列表中的節點進行翻轉。示例：輸入：(7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4) 輸出：7 -> 8 -> 0 -> 7 ``` ``` ## 前置知識 - 鏈表 - 棧 ## 公司 - 騰訊 - 百度 - 字節 ## 思路由于需要從低位開始加，然后進位。因此可以采用棧來簡化操作。依次將兩個鏈表的值分別入棧 stack1 和 stack2，然后相加入棧 stack，進位操作用一個變量 carried 記錄即可。最后根據 stack 生成最終的鏈表即可。 > 也可以先將兩個鏈表逆置，然后相加，最后將結果再次逆置。 ## 關鍵點解析 - 棧的基本操作 - carried 變量記錄進位 - 循環的終止條件設置成`stack.length > 0` 可以簡化操作 - 注意特殊情況，比如 1 + 99 = 100 ## 代碼 - 語言支持：JS，C++, Python3 JavaScript Code: ``` <pre class="calibre18">``` /* * @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： ``` <pre class="calibre18">``` /** * 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: // 此處若返回而非傳入vector，跑完所有測試用例多花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。跑完所有測試用例比第一種解法少花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: ``` <pre class="calibre18">``` # 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 ``` ``` **復雜度分析** 其中 M 和 N 分別為兩個鏈表的長度。 - 時間復雜度：O(M+N)O(M + N)O(M+N) - 空間復雜度：O(M+N)O(M + N)O(M+N) 大家對此有何看法，歡迎給我留言，我有時間都會一一查看回答。更多算法套路可以訪問我的 LeetCode 題解倉庫：<https://github.com/azl397985856/leetcode> 。目前已經 37K star 啦。大家也可以關注我的公眾號《力扣加加》帶你啃下算法這塊硬骨頭。 ![](https://img.kancloud.cn/cf/0f/cf0fc0dd21e94b443dd8bca6cc15b34b_900x500.jpg)