Leetcode | Multiply Strings

Given two numbers represented as strings, return multiplication of the numbers as a string.

Note: The numbers can be arbitrarily large and are non-negative.

还是不要想得太复杂。一开始还想着要4位4位二分去做。其实只要一位一位地乘，小心地维护好进位就好。

class Solution {
public:
    string multiply(string num1, string num2) {
        if (num1.empty()) return "";
        if (num2.empty()) return "";
        int m1, m2, carry, val;
        
        reverse(num1.begin(), num1.end());
        reverse(num2.begin(), num2.end());
        string ret(num1.length() + num2.length(), '0');
        for (int i = 0; i < num2.length(); ++i) {
            m2 = num2[i] - '0';
            carry = 0;
            for (int j = 0; j < num1.length(); ++j) {
                m1 = num1[j] - '0';
                val = m2 * m1 + carry + ret[i + j] - '0';
                ret[i + j] = val % 10 + '0';
                carry = val / 10;
            }
            ret[i + num1.length()] += carry;
        }

        string r = "0";
        for (int i = ret.length() - 1; i >= 0; --i) {
            if (ret[i] != '0') {
                r = ret.substr(0, i + 1);
                break;
            }
        }
        reverse(r.begin(), r.end());
        return r;
    }
};

reverse主要是便于取下标。

第10行，注意结果最大长度就是两个串的长度和；

第16行，注意每个位置需要加上ret[i+j]-'0'; 第17行，直接等于新的余数；

第20行，记得要把最后的进位保存下来；

第23行，如果从后往前扫一直没找到非零的数，那么证明结果为0，所以默认值设为'0'；

感觉第三次写比较长， 但是代码清晰许多。

class Solution {
public:
    string mul(string &num1, int n) {
        if (n == 0 || num1 == "0") return "0";
        if (n == 1) return num1;
        string ans = "";
        int carry = 0, v;
        for (int i = num1.length() - 1; i >= 0; --i) {
            v = carry + (num1[i] - '0') * n;
            ans.push_back((v % 10) + '0');
            carry = v / 10;
        }
        if (carry > 0) ans.push_back(carry + '0');
        reverse(ans.begin(), ans.end());
        return ans;
    }
    
    void add(string &ans, string &num) {
        int n1 = ans.length(), n2 = num.length(), v = 0, carry = 0;
        string tmp = "";
        for (int i = n1 - 1, j = n2 - 1; i >= 0 || j >= 0; --i, --j) {
            v = carry;
            if (i >= 0) v += (ans[i] - '0');
            if (j >= 0) v += (num[j] - '0');
            tmp.push_back((v % 10) + '0');
            carry = v / 10;
        }
        if (carry > 0) tmp.push_back(carry + '0');
        reverse(tmp.begin(), tmp.end());
        ans = tmp;
    }
    
    string multiply(string num1, string num2) {
         if (num1.empty() || num2.empty()) return "";
         string ans = "0";
         for (int i = num2.length() - 1; i >= 0; --i) {
             string sum = mul(num1, num2[i] - '0');
             if (num2.length() - i - 1 > 0 && sum != "0") sum.append(num2.length() - i - 1, '0');
             add(ans, sum);
         }
         return ans;
    }
};