415.字符串相加

链接：415.字符串相加
难度：Easy
标签：数学、字符串、模拟
简介：给定两个字符串形式的非负整数 num1 和 num2 ，计算它们的和。

题解 1 - typescript

• 编辑时间：2020-08-03
• 执行用时：92ms
• 内存消耗：40.2MB
• 编程语言：typescript
• 解法介绍：每个字符进行累加。

function addStrings(num1: string, num2: string): string {
  const arr1 = num1.split('').reverse();
  const arr2 = num2.split('').reverse();
  const ans: number[] = [];
  let go = 0;
  while (arr1.length !== 0 && arr2.length !== 0) {
    const num1 = arr1.shift() as string;
    const num2 = arr2.shift() as string;
    let num = parseInt(num1) + parseInt(num2) + go;
    if (num >= 10) {
      go = 1;
      num -= 10;
    } else go = 0;
    ans.push(num);
  }
  while (arr1.length !== 0) {
    let num = parseInt(arr1.shift()!) + go;
    if (num >= 10) {
      go = 1;
      num -= 10;
    } else go = 0;
    ans.push(num);
  }
  while (arr2.length !== 0) {
    let num = parseInt(arr2.shift()!) + go;
    if (num >= 10) {
      go = 1;
      num -= 10;
    } else go = 0;
    ans.push(num);
  }
  if (go === 1) ans.push(1);
  while (ans.length > 1 && ans[ans.length - 1] === 0) ans.pop();
  return ans.reverse().join('');
}

题解 2 - cpp

• 编辑时间：2023-07-17
• 执行用时：8ms
• 内存消耗：54.4MB
• 编程语言：cpp
• 解法介绍：遍历。

class Solution {
public:
    string addStrings(string num1, string num2) {
        if (num1.size() < num2.size()) swap(num1, num2);
        string res = "";
        reverse(num1.begin(), num1.end());
        reverse(num2.begin(), num2.end());
        int i = 0, add = 0;
        while (i < num1.size() || i < num2.size()) {
            int num = num1[i] - '0' + add;
            if (i < num2.size()) num += num2[i] - '0';
            if (num >= 10) {
                num -= 10;
                add = 1;
            } else {
                add = 0;
            }
            res = to_string(num) + res;
            i++;
        }
        if (add) res = "1" + res;
        return res;
    }
};

题解 3 - python

• 编辑时间：2023-07-17
• 执行用时：56ms
• 内存消耗：15.9MB
• 编程语言：python
• 解法介绍：同上。

class Solution:
    def addStrings(self, s1: str, s2: str) -> str:
        if len(s1) < len(s2):
            s1, s2 = s2, s1
        res = ""
        num1, num2 = list(s1), list(s2)
        num1.reverse()
        num2.reverse()
        i = add = 0
        while i < len(num1) or i < len(num2):
            num = ord(num1[i]) - ord('0') + add
            if i < len(num2):
                num += ord(num2[i]) - ord('0')
            if num >= 10:
                num -= 10
                add = 1
            else:
                add = 0
            res = str(num) + res
            i += 1
        if add:
            res = "1" + res
        return res

题解 4 - rust

• 编辑时间：2023-07-17
• 内存消耗：2.1MB
• 编程语言：rust
• 解法介绍：同上。

pub fn str_to_vec(s: &String) -> Vec<char> {
    s.chars().collect()
}
impl Solution {
    pub fn add_strings(num1: String, num2: String) -> String {
        let mut num1 = str_to_vec(&num1);
        let mut num2 = str_to_vec(&num2);
        num1.reverse();
        num2.reverse();
        if num1.len() < num2.len() {
            std::mem::swap(&mut num1, &mut num2);
        }
        let mut res = vec![];
        let mut i = 0;
        let mut add = 0;
        while i < num1.len() || i < num2.len() {
            let mut num = num1[i].to_digit(10).unwrap() as u8 + add;
            if i < num2.len() {
                num += num2[i].to_digit(10).unwrap() as u8;
            }
            if num >= 10 {
                num -= 10;
                add = 1;
            } else {
                add = 0;
            }
            res.push(num + b'0');
            i += 1;
        }
        if add != 0 {
            res.push(b'1');
        }
        res.reverse();
        String::from_utf8(res).unwrap()
    }
}