力扣题-1.3

[力扣刷题攻略] Re：从零开始的力扣刷题生活

力扣题1：43. 字符串相乘

解题思想：类似计算时采用的竖式乘法。首先取得num2的低位，并补齐对应的0，然后与num1进行相乘，然后进行字符串的相加操作。

class Solution(object):def multiply(self, num1, num2):""":type num1: str:type num2: str:rtype: str"""## 如果其中有个为0,则相乘后得0if num1 == "0" or num2 == "0":return "0"## 定义ans作为最后的返回值ans = "0"m = len(num1)n = len(num2)## 从低位开始遍历num2,用一位与num1进行乘法for i in range(n-1,-1,-1):add = 0y = int(num2[i])curr = ['0']*(n-1-i)## 与num1进行乘法操作for j in range(m-1,-1,-1):x = int(num1[j])temp = x*y+addcurr.append(str(temp%10))add = temp // 10if add > 0:curr.append(str(add))## 因为是通过append添加的最后需要进行翻转curr = "".join(curr[::-1])## 字符串的加法操作ans = self.addStrings(ans,curr)return ansdef addStrings(self, num1, num2):## 进行字符串的加法操作，从低位开始相加i, j = len(num1) - 1, len(num2) - 1add = 0ans = list()while i >= 0 or j >= 0 or add != 0:x = int(num1[i]) if i >= 0 else 0y = int(num2[j]) if j >= 0 else 0result = x + y + addans.append(str(result % 10))add = result // 10i -= 1j -= 1## 最后需要进行翻转return "".join(ans[::-1])

class Solution {
public:string multiply(string num1, string num2) {if (num1 == "0" || num2 == "0") {return "0";}string ans = "0";int m = num1.length();int n = num2.length();for (int i = n - 1; i >= 0; i--) {int add = 0;int y = num2[i] - '0';string curr(n - 1 - i, '0');for (int j = m - 1; j >= 0; j--) {int x = num1[j] - '0';int temp = x * y + add;curr.push_back(temp % 10 + '0');add = temp / 10;}if (add > 0) {curr.push_back(add + '0');}reverse(curr.begin(), curr.end());ans = addStrings(ans, curr);}return ans;}string addStrings(string num1, string num2) {int i = num1.length() - 1;int j = num2.length() - 1;int add = 0;string result;while (i >= 0 || j >= 0 || add != 0) {int x = (i >= 0) ? num1[i] - '0' : 0;int y = (j >= 0) ? num2[j] - '0' : 0;int sum = x + y + add;result.push_back(sum % 10 + '0');add = sum / 10;i--;j--;}reverse(result.begin(), result.end());return result;}
};