【笔记】语言实例比较 2. 两数之和 C++ Rust Java Python

本文介绍: 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。输入：l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]输入：l1 = [2,4,3], l2 = [5,6,4]请你将两个数相加，并以相同形式返回一个表示和的链表。语言实例比较 2. 两数之和 C++ Rust。每个链表中的节点数在范围 [1, 100] 内。输入：l1 = [0], l2 = [0]输出：[8,9,9,9,0,0,0,1]输出：[7,0,8]

语言实例比较 2. 两数之和 C++ Rust Java Python

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：
输入：l1 = [2,4,3], l2 = [5,6,4]
输出：[7,0,8]
解释：342 + 465 = 807.

示例 2：
输入：l1 = [0], l2 = [0]
输出：[0]

/**
 * 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 *dummyHead = new ListNode();
        ListNode *cur = dummyHead;
        int sum = 0;
        while(l1 || l2 || sum){
            if(l1) {
                sum += l1->val;
                l1 = l1->next;
            }
            if(l2) {
                sum += l2->val;
                l2 = l2->next;
            }
            cur->next = new ListNode(sum % 10, nullptr);
            cur = cur->next;
            sum = sum / 10;
        }

        return dummyHead->next;
    }
};

// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
    pub fn add_two_numbers(l1: Option<Box<ListNode>>, l2: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
        carried(l1, l2, 0)
    }
}

pub fn carried(l1: Option<Box<ListNode>>, l2: Option<Box<ListNode>>, mut carry: i32) -> Option<Box<ListNode>> {
    if l1.is_none() && l2.is_none() && carry == 0 {
        None
    } else {
        Some(Box::new(ListNode {
            next: carried(
                l1.and_then(|x| {carry += x.val; x.next}), 
                l2.and_then(|x| {carry += x.val; x.next}), 
                carry / 10
            ),
            val: carry % 10
        }))
    }
}

// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
    pub fn add_two_numbers(l1: Option<Box<ListNode>>, l2: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
        let mut l1 = l1;
        let mut l2 = l2;
        
        let mut dummy = ListNode::new(0); // 哨兵节点
        let mut cur = &mut dummy;
        let mut sum = 0; // 剩余和，进位
        while l1.is_some() || l2.is_some() || sum != 0 {
            l1 = l1.and_then(|node| {sum += node.val; node.next});
            l2 = l2.and_then(|node| {sum += node.val; node.next});
            cur.next = Some(Box::new(ListNode::new(sum % 10))); // 每个节点保存一个数位
            cur = cur.next.as_mut().unwrap(); // 下一个节点
            sum = sum / 10; // 新的进位
        }
        dummy.next // 哨兵节点的下一个节点就是头节点
    }
}

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode();
        ListNode cur = dummy;
        int sum = 0;
        while(l1 != null || l2 != null || sum != 0) {
            if (l1 != null) { sum += l1.val; l1 = l1.next; }
            if (l2 != null) { sum += l2.val; l2 = l2.next; }
            cur.next = new ListNode(sum % 10);
            cur = cur.next;
            sum /= 10;
        }
        return dummy.next;
    }
}

# 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: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        cur = dummy = ListNode()
        sum = 0
        while l1 or l2 or sum:
            if l1: l1, sum = l1.next, sum + l1.val
            if l2: l2, sum = l2.next, sum + l2.val
            cur.next = ListNode(sum % 10)
            cur = cur.next
            sum //= 10
        return dummy.next