Leetcode 43 Multiply Strings

关注

原创最新推荐文章于 2026-04-07 10:22:05 发布 769 阅读 AI 写同款· GEO 优化 ›

43. Multiply Strings [Leetcode] [FFT] [python] 查看题目一道大数相乘的题，大数也可以看成是多项式，再运用FFT来处理。粗略步骤： 1. 将多项式从系数表示转为点值表示 2. 把两个多项式的点值表示相乘 3. 将相乘后的点值表示转为系数表示 1. 将多项式从系数表示转为点值表示系数表示： A(x)=a0+a1x+a2x2+⋅⋅⋅+an−1xn−1A(x)=a0+a1x+a2x2+···+an−1xn−1... 阅读详情

class Solution:
    # @param {string} num1
    # @param {string} num2
    # @return {string}
    def multiply(self, num1, num2):
        if num1 == "0" or num2 == "0":      return "0"
        
        result = [0] * (len(num1) + len(num2))
        num1 = [int(i) for i in num1]
        num2 = [int(i) for i in num2]
        
        for index1 in xrange(len(num1)):
            multiplier = num1[index1]
            temp = [i*multiplier for i in num2]         # Multiply
            temp.extend([0] * (len(num1) - index1 - 1)) # Shift
            
            for resIndex in xrange(1, len(temp) + 1):
                result[-resIndex] += temp[-resIndex]
        
        for resIndex in xrange(len(result)-1, 0, -1):
            result[resIndex-1] += result[resIndex] // 10
            result[resIndex] %= 10
        
        result = "".join([str(i) for i in result]).lstrip("0")        
        return result

class Solution:
    # @param num1, a string
    # @param num2, a string
    # @return a string
    def multiply(self, num1, num2):
        num1 = num1[::-1]; num2 = num2[::-1]
        arr = [0 for i in range(len(num1)+len(num2))]
        for i in range(len(num1)):
            for j in range(len(num2)):
                arr[i+j] += int(num1[i]) * int(num2[j])
        ans = []
        for i in range(len(arr)):
            digit = arr[i] % 10
            carry = arr[i] / 10
            if i < len(arr)-1:
                arr[i+1] += carry
            ans.insert(0, str(digit))
        while ans[0] == '0' and len(ans) > 1:
            del ans[0]
        return ''.join(ans)