README.md

comments	difficulty	edit_url	tags
true	中等	https://github.com/doocs/leetcode/edit/main/solution/0400-0499/0445.Add%20Two%20Numbers%20II/README.md	栈 链表 数学

445. 两数相加 II

English Version

题目描述

给你两个 非空 链表来代表两个非负整数。数字最高位位于链表开始位置。它们的每个节点只存储一位数字。将这两数相加会返回一个新的链表。

你可以假设除了数字 0 之外，这两个数字都不会以零开头。

 

示例1：

输入：l1 = [7,2,4,3], l2 = [5,6,4]
输出：[7,8,0,7]

示例2：

输入：l1 = [2,4,3], l2 = [5,6,4]
输出：[8,0,7]

示例3：

输入：l1 = [0], l2 = [0]
输出：[0]

 

提示：

• 链表的长度范围为 [1, 100]
• 0 <= node.val <= 9
• 输入数据保证链表代表的数字无前导 0

 

进阶：如果输入链表不能翻转该如何解决？

解法

方法一：翻转

手动翻转链表 l1 与 l2，将此题转换为 2. 两数相加，相加过程一致。对于最后返回的结果链表也需要进行翻转，共计三次。

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]:
        s1, s2 = [], []
        while l1:
            s1.append(l1.val)
            l1 = l1.next
        while l2:
            s2.append(l2.val)
            l2 = l2.next
        dummy = ListNode()
        carry = 0
        while s1 or s2 or carry:
            s = (0 if not s1 else s1.pop()) + (0 if not s2 else s2.pop()) + carry
            carry, val = divmod(s, 10)
            # node = ListNode(val, dummy.next)
            # dummy.next = node
            dummy.next = ListNode(val, dummy.next)
        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) {
        Deque<Integer> s1 = new ArrayDeque<>();
        Deque<Integer> s2 = new ArrayDeque<>();
        for (; l1 != null; l1 = l1.next) {
            s1.push(l1.val);
        }
        for (; l2 != null; l2 = l2.next) {
            s2.push(l2.val);
        }
        ListNode dummy = new ListNode();
        int carry = 0;
        while (!s1.isEmpty() || !s2.isEmpty() || carry != 0) {
            int s = (s1.isEmpty() ? 0 : s1.pop()) + (s2.isEmpty() ? 0 : s2.pop()) + carry;
            // ListNode node = new ListNode(s % 10, dummy.next);
            // dummy.next = node;
            dummy.next = new ListNode(s % 10, dummy.next);
            carry = s / 10;
        }
        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) {
        stack<int> s1;
        stack<int> s2;
        for (; l1; l1 = l1->next) s1.push(l1->val);
        for (; l2; l2 = l2->next) s2.push(l2->val);
        ListNode* dummy = new ListNode();
        int carry = 0;
        while (!s1.empty() || !s2.empty() || carry) {
            int s = carry;
            if (!s1.empty()) {
                s += s1.top();
                s1.pop();
            }
            if (!s2.empty()) {
                s += s2.top();
                s2.pop();
            }
            // ListNode* node = new ListNode(s % 10, dummy->next);
            // dummy->next = node;
            dummy->next = new ListNode(s % 10, dummy->next);
            carry = s / 10;
        }
        return dummy->next;
    }
};

Go

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
	s1, s2 := arraystack.New(), arraystack.New()
	for l1 != nil {
		s1.Push(l1.Val)
		l1 = l1.Next
	}
	for l2 != nil {
		s2.Push(l2.Val)
		l2 = l2.Next
	}
	carry, dummy := 0, new(ListNode)
	for !s1.Empty() || !s2.Empty() || carry > 0 {
		s := carry
		v, ok := s1.Pop()
		if ok {
			s += v.(int)
		}
		v, ok = s2.Pop()
		if ok {
			s += v.(int)
		}
		// node := &ListNode{s % 10, dummy.Next}
		// dummy.Next = node
		dummy.Next = &ListNode{s % 10, dummy.Next}
		carry = s / 10
	}
	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 s1: number[] = [];
    const s2: number[] = [];
    for (; l1; l1 = l1.next) {
        s1.push(l1.val);
    }
    for (; l2; l2 = l2.next) {
        s2.push(l2.val);
    }
    const dummy = new ListNode();
    let carry = 0;
    while (s1.length || s2.length || carry) {
        const s = (s1.pop() ?? 0) + (s2.pop() ?? 0) + carry;
        // const node = new ListNode(s % 10, dummy.next);
        // dummy.next = node;
        dummy.next = new ListNode(s % 10, dummy.next);
        carry = Math.floor(s / 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 {
    fn reverse(mut head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
        let mut pre = None;
        while let Some(mut node) = head {
            let next = node.next.take();
            node.next = pre.take();
            pre = Some(node);
            head = next;
        }
        pre
    }

    pub fn add_two_numbers(
        mut l1: Option<Box<ListNode>>,
        mut l2: Option<Box<ListNode>>,
    ) -> Option<Box<ListNode>> {
        l1 = Self::reverse(l1);
        l2 = Self::reverse(l2);
        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;
        }
        Self::reverse(dummy.unwrap().next.take())
    }
}

方法二：栈

我们可以使用两个栈 $s1$ 和 $s2$ 分别存储两个链表元素，然后同时遍历两个栈，并使用变量 $carry$ 表示当前是否有进位。

遍历时，我们弹出对应栈的栈顶元素，计算它们与进位 $carry$ 的和，然后更新进位的值，并创建一个链表节点，插入答案链表的头部。如果两个栈都遍历结束，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 $O(\max(m, n))$，空间复杂度 $O(m + n)$。其中 $m$ 和 $n$ 分别为两个链表的长度。

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 {
    fn create_stack(mut head: Option<Box<ListNode>>) -> Vec<i32> {
        let mut res = vec![];
        while let Some(node) = head {
            res.push(node.val);
            head = node.next;
        }
        res
    }

    pub fn add_two_numbers(
        l1: Option<Box<ListNode>>,
        l2: Option<Box<ListNode>>,
    ) -> Option<Box<ListNode>> {
        let mut s1 = Self::create_stack(l1);
        let mut s2 = Self::create_stack(l2);

        let mut dummy = Box::new(ListNode::new(0));
        let mut carry = 0;
        while !s1.is_empty() || !s2.is_empty() || carry != 0 {
            if let Some(val) = s1.pop() {
                carry += val;
            }
            if let Some(val) = s2.pop() {
                carry += val;
            }
            dummy.next = Some(Box::new(ListNode {
                val: carry % 10,
                next: dummy.next.take(),
            }));
            carry /= 10;
        }
        dummy.next.take()
    }
}