字符串加减乘除

大数加减乘除运算总结 - 力扣（LeetCode）

加减乘除

大数相加 -415 easy
注意通过数组记录各个字符串, 同时也有注意用 carry 来记录进位, 注意 reversed 方法.

class Solution:
    def addStrings(self, num1: str, num2: str) -> str:
        ptr_num1 = 0
        ptr_num2 = 0
        return_num = []
        carry = 0
        while ptr_num1 < len(num1) or ptr_num2 < len(num2) or carry == 1:
            digit_1 = int(num1[len(num1) - 1 - ptr_num1]) if ptr_num1 < len(num1) else 0
            digit_2 = int(num2[len(num2) - 1 - ptr_num2]) if ptr_num2 < len(num2) else 0
            curr_digit = carry + digit_1 + digit_2
            carry = 0 if curr_digit < 10 else 1
            return_num.append(str(curr_digit % 10))
            ptr_num2 += 1
            ptr_num1 += 1
        reversed_num = reversed(return_num)
        return str("".join(reversed_num))

大数相互乘法 -43 mid

1. 构造一个长度为 len(num1) + len(num2) 来放 value 数组
2. 注意要去除多余的 0, 如 0011 前的两个 0, 同时还要考虑 0*N 这样的情况
3. 分两个循环, 一个制作 value 数组, 一个处理 value 数组

class Solution:
    def multiply(self, num1: str, num2: str) -> str:
        ptr_num1 = 0
        carry = 0
        return_arr = []
        value = [0 for _ in range(len(num1) + len(num2))]
        while ptr_num1 < len(num1):
            digit_1 = int(num1[len(num1) - ptr_num1 - 1])
            for i in range(0, len(num2)):
                digit_2 = int(num2[len(num2) - i - 1])
                curr_digit = digit_1 * digit_2
                value[-(i + 1 + ptr_num1)] += curr_digit
            ptr_num1 += 1
        for i in range(len(value)):
            curr = len(value) - 1 - i
            curr_value = value[curr] + carry
            return_arr.append(str(curr_value % 10))
            carry = curr_value // 10
        index = 0
        number_str = "".join(reversed(return_arr[index::]))
        for i in range(len(number_str)):
            if number_str[i] == "0":
                index += 1
            else:
                break
        if index == len(num1) + len(num2):
            return "0"
        return number_str[index::]

这个看着要清晰的多

function multiply(num1: string, num2: string): string {
   if(num1==="0"||num2==="0"){
      return "0"
   }
   let stack = [[]];
   let num1Arr = num1.split("");
   let num2Arr = num2.split("");
   num1Arr = num1Arr.reverse();
   num2Arr = num2Arr.reverse();
   for(let i = 0;i<num1Arr.length;i++){
      for(let j=0;j<num2Arr.length;j++){
         if(i+j>=stack.length){
            stack.push([]);
         }
         stack[i+j].push(parseInt(num1Arr[i])*parseInt(num2Arr[j]))
      }
   }
   let sum = "";
   for (let i = 0;i < stack.length-1;i++){
      let curSum = stack[i].reduce((prev,curr)=>prev+curr,0);
      stack[i] = [curSum%10]
      stack[i+1].push(Math.floor(curSum/10));
   }
   stack.reverse();
   for(let i=0;i<stack.length;i++){
      sum += (stack[i].reduce((prev,curr)=>prev+curr,0)).toString();
   }
   return sum
}```


大数减法,  
同加法, 但是要考虑出现负数的情况, 在计算前要先比较大小.

同样类型的还有非 10 进制的处理, 这里在 `curr_digit = carry + digit_1 + digit_2` 处理就好,  
同时注意写两个转换函数  
[Python 二进制，十进制，十六进制转换\_python二进制转10进制-CSDN博客](https://blog.csdn.net/u012063703/article/details/42609833)