Leetcode 445. 两数相加 II

news2024/11/25 0:54:58

原题链接：Leetcode445. Add Two Numbers II

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example 1:
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Input: l1 = [7,2,4,3], l2 = [5,6,4]
Output: [7,8,0,7]

Example 2:

Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [8,0,7]

Example 3:

Input: l1 = [0], l2 = [0]
Output: [0]

Constraints:

The number of nodes in each linked list is in the range [1, 100].
0 <= Node.val <= 9
It is guaranteed that the list represents a number that does not have leading zeros.

Follow up: Could you solve it without reversing the input lists?

方法一：栈

思路：

两个数是在链表中逆序存储的，而计算的时候需要把顺序倒回来，因此就想到栈。
其余的部分正常模拟即可

C++代码：

/**
 * 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) {

        stack<int> st1, st2;    // 设置两个栈用于逆序存数
        int carry = 0;  // 进位
        ListNode* ans = nullptr; 

        // 将两个数存入栈中
        while(l1){
            st1.push(l1->val);
            l1 = l1->next;
        }
        while(l2){
            st2.push(l2->val);
            l2 = l2->next;
        }

        while(!st1.empty() || !st2.empty() || carry > 0){
            int a = st1.empty() ? 0 : st1.top();
            int b = st2.empty() ? 0 : st2.top();

            if(!st1.empty()) st1.pop();
            if(!st2.empty()) st2.pop();

            int cur = a + b + carry;
            
            carry = cur / 10;
            cur = cur % 10;

            // 创建一个新节点
            auto cur_node = new ListNode(cur);

            cur_node -> next = ans;
            ans = cur_node;
        }
        return ans;
    }
};