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中等
递归
链表
数学

2. 两数相加

English Version

题目描述

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]

示例 3：

输入：l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
输出：[8,9,9,9,0,0,0,1]

 

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

解法

方法一：模拟

我们同时遍历两个链表 $l_1$ 和 $l_2$，并使用变量 $carry$ 表示当前是否有进位。

每次遍历时，我们取出对应链表的当前位，计算它们与进位 $carry$ 的和，然后更新进位的值，最后将当前位的值加入答案链表。如果两个链表都遍历完了，并且进位为 $0$ 时，遍历结束。

最后我们返回答案链表的头节点即可。

时间复杂度 $O(\max(m, n))$，其中 $m$ 和 $n$ 分别为两个链表的长度。我们需要遍历两个链表的全部位置，而处理每个位置只需要 $O(1)$ 的时间。忽略答案的空间消耗，空间复杂度 $O(1)$。

Python3

# 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]:
        dummy = ListNode()
        carry, curr = 0, dummy
        while l1 or l2 or carry:
            s = (l1.val if l1 else 0) + (l2.val if l2 else 0) + carry
            carry, val = divmod(s, 10)
            curr.next = ListNode(val)
            curr = curr.next
            l1 = l1.next if l1 else None
            l2 = l2.next if l2 else None
        return dummy.next

Java

/**
 * 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(0);
        int carry = 0;
        ListNode cur = dummy;
        while (l1 != null || l2 != null || carry != 0) {
            int s = (l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + carry;
            carry = s / 10;
            cur.next = new ListNode(s % 10);
            cur = cur.next;
            l1 = l1 == null ? null : l1.next;
            l2 = l2 == null ? null : l2.next;
        }
        return dummy.next;
    }
}

C++

/**
 * 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* dummy = new ListNode();
        int carry = 0;
        ListNode* cur = dummy;
        while (l1 || l2 || carry) {
            int s = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + carry;
            carry = s / 10;
            cur->next = new ListNode(s % 10);
            cur = cur->next;
            l1 = l1 ? l1->next : nullptr;
            l2 = l2 ? l2->next : nullptr;
        }
        return dummy->next;
    }
};

Go

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
	dummy := &ListNode{}
	carry := 0
	cur := dummy
	for l1 != nil || l2 != nil || carry != 0 {
		s := carry
		if l1 != nil {
			s += l1.Val
		}
		if l2 != nil {
			s += l2.Val
		}
		carry = s / 10
		cur.Next = &ListNode{s % 10, nil}
		cur = cur.Next
		if l1 != nil {
			l1 = l1.Next
		}
		if l2 != nil {
			l2 = l2.Next
		}
	}
	return dummy.Next
}

TypeScript

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function addTwoNumbers(l1: ListNode | null, l2: ListNode | null): ListNode | null {
    const dummy = new ListNode();
    let cur = dummy;
    let 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 = Math.floor(sum / 10);
    }
    return dummy.next;
}

Rust

// 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(
        mut l1: Option<Box<ListNode>>,
        mut l2: Option<Box<ListNode>>,
    ) -> Option<Box<ListNode>> {
        let mut dummy = Some(Box::new(ListNode::new(0)));
        let mut cur = &mut dummy;
        let mut sum = 0;
        while l1.is_some() || l2.is_some() || sum != 0 {
            if let Some(node) = l1 {
                sum += node.val;
                l1 = node.next;
            }
            if let Some(node) = l2 {
                sum += node.val;
                l2 = node.next;
            }
            cur.as_mut().unwrap().next = Some(Box::new(ListNode::new(sum % 10)));
            cur = &mut cur.as_mut().unwrap().next;
            sum /= 10;
        }
        dummy.unwrap().next.take()
    }
}

JavaScript

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var addTwoNumbers = function (l1, l2) {
    const dummy = new ListNode();
    let carry = 0;
    let cur = dummy;
    while (l1 || l2 || carry) {
        const s = (l1?.val || 0) + (l2?.val || 0) + carry;
        carry = Math.floor(s / 10);
        cur.next = new ListNode(s % 10);
        cur = cur.next;
        l1 = l1?.next;
        l2 = l2?.next;
    }
    return dummy.next;
};

C#

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

PHP

/**
 * Definition for a singly-linked list.
 * class ListNode {
 *     public $val = 0;
 *     public $next = null;
 *     function __construct($val = 0, $next = null) {
 *         $this->val = $val;
 *         $this->next = $next;
 *     }
 * }
 */
class Solution {
    /**
     * @param ListNode $l1
     * @param ListNode $l2
     * @return ListNode
     */
    function addTwoNumbers($l1, $l2) {
        $dummy = new ListNode(0);
        $current = $dummy;
        $carry = 0;

        while ($l1 !== null || $l2 !== null) {
            $x = $l1 !== null ? $l1->val : 0;
            $y = $l2 !== null ? $l2->val : 0;

            $sum = $x + $y + $carry;
            $carry = (int) ($sum / 10);
            $current->next = new ListNode($sum % 10);
            $current = $current->next;

            if ($l1 !== null) {
                $l1 = $l1->next;
            }

            if ($l2 !== null) {
                $l2 = $l2->next;
            }
        }

        if ($carry > 0) {
            $current->next = new ListNode($carry);
        }

        return $dummy->next;
    }
}

Swift

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init() { self.val = 0; self.next = nil; }
 *     public init(_ val: Int) { self.val = val; self.next = nil; }
 *     public init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next; }
 * }
 */
class Solution {
    func addTwoNumbers(_ l1: ListNode?, _ l2: ListNode?) -> ListNode? {
        var dummy = ListNode.init()
        var carry = 0
        var l1 = l1
        var l2 = l2
        var cur = dummy
        while l1 != nil || l2 != nil || carry != 0 {
            let s = (l1?.val ?? 0) + (l2?.val ?? 0) + carry
            carry = s / 10
            cur.next = ListNode.init(s % 10)
            cur = cur.next!
            l1 = l1?.next
            l2 = l2?.next
        }
        return dummy.next
    }
}

Ruby

# Definition for singly-linked list.
# class ListNode
#     attr_accessor :val, :next
#     def initialize(val = 0, _next = nil)
#         @val = val
#         @next = _next
#     end
# end
# @param {ListNode} l1
# @param {ListNode} l2
# @return {ListNode}
def add_two_numbers(l1, l2)
    dummy = ListNode.new()
    carry = 0
    cur = dummy
    while !l1.nil? || !l2.nil? || carry > 0
        s = (l1.nil? ? 0 : l1.val) + (l2.nil? ? 0 : l2.val) + carry
        carry = s / 10
        cur.next = ListNode.new(s % 10)
        cur = cur.next
        l1 = l1.nil? ? l1 : l1.next
        l2 = l2.nil? ? l2 : l2.next
    end
    dummy.next
end

Nim

#[
    # Driver code in the solution file
    # Definition for singly-linked list.
    type
    Node[int] = ref object
        value: int
        next: Node[int]

    SinglyLinkedList[T] = object
        head, tail: Node[T]
]#

# More efficient code churning ...
proc addTwoNumbers(l1: var SinglyLinkedList, l2: var SinglyLinkedList): SinglyLinkedList[int] =
  var
    aggregate: SinglyLinkedList
    psum: seq[char]
    temp_la, temp_lb: seq[int]

  while not l1.head.isNil:
    temp_la.add(l1.head.value)
    l1.head = l1.head.next

  while not l2.head.isNil:
    temp_lb.add(l2.head.value)
    l2.head = l2.head.next

  psum = reversed($(reversed(temp_la).join("").parseInt() + reversed(temp_lb).join("").parseInt()))
  for i in psum: aggregate.append(($i).parseInt())

  result = aggregate