两数相加 II(链表版)

题目

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. 这两个数字都不会以零开头.

示例

输入: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)

输出: 7 -> 8 -> 0 -> 7

题解

这道题是 两数相加(链表版) 的进阶版, 难点在于高位在前, 低位在后, 故无法直接使用加法规则. 为了使低位先出来, 自然会想到使用栈.

1. 遍历两个链表, 将数据逐一 push 到栈中
2. 对两个栈进行 pop, 得到的数字即为相同位置的数字
3. 剩下的就是大数相加和栈的基本操作了

JavaScript

/**
 * @param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}
 */
var addTwoNumbers = function (l1, l2) {
  const stack1 = []
  const stack2 = []

  while (l1 !== null) {
    stack1.push(l1.val)
    l1 = l1.next
  }

  while (l2 !== null) {
    stack2.push(l2.val)
    l2 = l2.next
  }

  const head = new ListNode(0)
  let carry = 0

  while (stack1.length !== 0 || stack2.length !== 0 || carry !== 0) {
    const val1 = stack1.length === 0 ? 0 : stack1.pop()
    const val2 = stack2.length === 0 ? 0 : stack2.pop()
    const sum = val1 + val2 + carry
    carry = (sum / 10) | 0

    const node = new ListNode(sum % 10)
    node.next = head.next
    head.next = node
  }

  return head.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 }
    }
}

pub fn add_two_numbers(
    l1: Option<Box<ListNode>>,
    l2: Option<Box<ListNode>>,
) -> Option<Box<ListNode>> {
    let mut a = Vec::new();
    let mut b = Vec::new();

    let mut ll = l1;
    while let Some(x) = ll {
        a.push(x.val);
        ll = x.next;
    }

    let mut ll = l2;
    while let Some(x) = ll {
        b.push(x.val);
        ll = x.next;
    }

    let mut carry = 0;
    let mut head = ListNode::new(0);
    while a.len() > 0 || b.len() > 0 || carry != 0 {
        let mut a_num = 0;
        let mut b_num = 0;
        if a.len() > 0 {
            a_num = a.pop().unwrap();
        }
        if b.len() > 0 {
            b_num = b.pop().unwrap();
        }

        let sum = a_num + b_num + carry;

        let mut node = ListNode::new(sum % 10);
        node.next = head.next.clone();
        head.next = Some(Box::new(node));

        carry = sum / 10;
    }

    head.next
}