2. 两数相加

题目

给你两个 非空 的链表，表示两个非负的整数。它们每位数字都是按照 逆序 的方式存储的，并且每个节点只能存储 一位 数字。

请你将两个数相加，并以相同形式返回一个表示和的链表。

你可以假设除了数字 0 之外，这两个数都不会以 0 开头。

示例 1：

示例 2：

示例 3：

提示：

• 每个链表中的节点数在范围 [1, 100] 内
• 0 <= Node.val <= 9
• 题目数据保证列表表示的数字不含前导零

解答（C++）

方法一

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        int flag = 0;
        ListNode* ans = new ListNode();
        ListNode* p = ans;
        while(l1 && l2) {
            int sumOfTwo = l1->val + l2->val + flag;
            ListNode* temp = new ListNode(sumOfTwo % 10);
            flag = sumOfTwo / 10;
            l1 = l1->next;
            l2 = l2->next;
            p->next = temp;
            p = p->next;
        }
        while(l1) {
            int sumOfTwo = l1->val + flag;
            ListNode* temp = new ListNode(sumOfTwo % 10);
            flag = sumOfTwo / 10;
            l1 = l1->next;
            p->next = temp;
            p = p->next;
        }
        while(l2) {
            int sumOfTwo = l2->val + flag;
            ListNode* temp = new ListNode(sumOfTwo % 10);
            flag = sumOfTwo / 10;
            l2 = l2->next;
            p->next = temp;
            p = p->next; 
        }
        if(flag) {
            p->next = new ListNode(1);
        }
        return ans->next;
    }
};

简化一下

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        int flag = 0;
        ListNode* ans = new ListNode();
        ListNode* p = ans;
        while(l1 || l2) {
            int num1 = l1? l1->val:0;
            int num2 = l2? l2->val:0;
            int sumOfTwo = num1 + num2 + flag;
            ListNode* temp = new ListNode(sumOfTwo % 10);
            flag = sumOfTwo / 10;
            l1 = l1? l1->next:nullptr;
            l2 = l2? l2->next:nullptr;
            p->next = temp;
            p = p->next;
        }
        if(flag) {
            p->next = new ListNode(1);
        }
        return ans->next;
    }
};

优化一下

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* head = nullptr;
        ListNode* tail = nullptr;
        int flag = 0;
        while(l1 || l2) {
            int num1 = l1? l1->val:0;
            int num2 = l2? l2->val:0;
            int sumOfTwo = num1 + num2 + flag;
            flag = sumOfTwo / 10;
            if(head == nullptr) {
                head = tail = new ListNode(sumOfTwo % 10);
            }else {
                tail->next = new ListNode(sumOfTwo % 10);
                tail = tail->next;
            }
            if(l1) l1 = l1->next;
            if(l2) l2 = l2->next;
        }
        if(flag) tail->next = new ListNode(1);
        return head;
    }
};

解答（Python）

方法一

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        head = tail = ListNode(0)
        flag = 0
        while l1 or l2:
            num1 = l1.val if l1 else 0
            num2 = l2.val if l2 else 0
            sumOfTwo = num1 + num2 + flag
            flag = sumOfTwo // 10
            tail.next = ListNode(sumOfTwo % 10)
            tail = tail.next
            if l1:
                l1 = l1.next
            if l2:
                l2 = l2.next
        if flag:
            tail.next = ListNode(1)
        return head.next

优化一下

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        head = tail = None
        flag = 0
        while l1 or l2:
            num1 = l1.val if l1 else 0
            num2 = l2.val if l2 else 0
            sumOfTwo = num1 + num2 + flag
            flag = sumOfTwo // 10
            if head is None:
                head = tail = ListNode(sumOfTwo % 10)
            else:
                tail.next = ListNode(sumOfTwo % 10)
                tail = tail.next
            if l1:
                l1 = l1.next
            if l2:
                l2 = l2.next
        if flag:
            tail.next = ListNode(1)
        return head