leetcode 43. Multiply Strings

对于python3来说，有两种解法。第一种因为有长整型，不用考虑整型的溢出，所以可以直接这么写：

# solution1 python有长整型，所以不用担心溢出
    def char_to_int(self, char):
        integers = {'0':0,'1':1,'2':2,'3':3,'4':4,'5':5,'6':6,'7':7,'8':8,'9':9}
        return integers[char]
      
    def int_to_char(self, integer):
        chars = {0:'0',1:'1',2:'2',3:'3',4:'4',5:'5',6:'6',7:'7',8:'8',9:'9'}
        return chars[integer]
        
    def multiply(self, num1: str, num2: str) -> str:
        # string 转 int
        level = 10**(len(num1)-1)
        num1_int = 0
        for num in num1:
            num1_int += self.char_to_int(num) * level
            level //= 10
        level = 10**(len(num2)-1)
        num2_int = 0
        for num in num2:
            num2_int += self.char_to_int(num) * level
            level //= 10
         
        num_ans_int = num1_int * num2_int

        # int 转 string
        num_ans = ""
        while num_ans_int/10 !=0:
            num_ans = self.int_to_char(num_ans_int%10) + num_ans
            num_ans_int = num_ans_int // 10
        if num_ans == "":
            num_ans = "0"
        return num_ans

第二种解法，假设python 没有长整型：

按照两数相乘的过程，来计算。计算可以分成两部分，用 123456*7654321 来举例：

分为计算红线左边和红线右边

    def multiply(self, num1:str, num2:str) ->str:
        # 计算红线右边
        num_ans = ""
        last = 0
        for i1 in range(1,len(num1)+1):
            m1 = len(num1) - i1
            temp = last
            for i2 in range(min(i1,len(num2))):
                m2 = len(num2) - i2 -1
                temp += self.char_to_int(num1[m1+i2]) * self.char_to_int(num2[m2])
            last = temp // 10
            num_ans = self.int_to_char(temp%10) + num_ans
        
        #计算红线左边
        for i2 in range(1,len(num2)):
            m2 = len(num2) - i2 -1
            temp = last
            for m1 in range(min(m2+1,len(num1))):
                temp+= self.char_to_int(num1[m1]) * self.char_to_int(num2[m2-m1])
            last = temp // 10
            num_ans = self.int_to_char(temp%10) + num_ans
        while last/10 !=0:
            num_ans = self.int_to_char(last%10) + num_ans
            last = last // 10
        
        # 一些边缘情况
        if num_ans =="":
            return "0"
        change = True
        for num in num_ans:
            if num == "0":
                pass
            else:
                change = False
                break
        if change:
            num_ans = "0"
        return num_ans