LeetCode 43 Multiply Strings

题意：

输入2个表示非负数的字符串，求它们的乘积。

思路：

暴力模拟吧！！19ms能过。

我利用单个位置上数字只有10个来优化乘法次数，效率能提高到9ms。

代码：

/** * 利用数字个数优化 9ms */class Solution {public:    string multiply(string num1, string num2) {        if (num1.size() < num2.size()) {            swap(num1, num2);        }        vector<int> ans{0};        vector<int> number1 = convert(num1), number2 = convert(num2);        counted[1] = number1;        for (int i = 0; i < number2.size(); ++i) {            if (number2[i] == 0) {                continue;            }            int b = number2[i];            if (counted[b].size() > 0) {                add(ans, counted[b], i);            } else {                vector<int> tmp = multiply(number1, b);                add(ans, tmp, i);                counted[b] = tmp;            }        }        return convert(ans);    }private:    vector<int> counted[10];    vector<int> convert(string num) {        vector<int> number;        for (int i = num.size() - 1; i >= 0; --i) {            number.push_back(num[i] - '0');        }        return number;    }    string convert(vector<int> num) {        stringstream ss;        bool zero = true;        for (int i = num.size() - 1; i >= 1; --i) {            if (zero && num[i] == 0) {                continue;            } else {                zero = false;            }            ss << num[i];        }        ss << num[0];        return ss.str();    }    vector<int> multiply(vector<int> &num, int b) {        vector<int> ans;        int tmp = 0;        for (int i = 0; i < num.size(); ++i) {            tmp += b * num[i];            ans.push_back(tmp % 10);            tmp /= 10;        }        if (tmp != 0) {            ans.push_back(tmp);        }        return ans;    }    void add(vector<int> &num1, vector<int> &num2, int pos) {        for (int i = 0; i < pos + 1 - (int)num1.size(); ++i) {            num1.push_back(0);        }        int tmp = 0;        for (int i = pos; i < num1.size() || i < num2.size() + pos; ++i) {            if (i < num2.size() + pos) {                if (i >= pos) {                    tmp += num2[i - pos];                }            } else {                if (tmp == 0) {                    break;                }            }            if (i < num1.size()) {                tmp += num1[i];            } else {                num1.push_back(0);            }            num1[i] = tmp % 10;            tmp /= 10;        }        if (tmp != 0) {            num1.push_back(tmp);        }    }};/** * string转成vector<int>直接计算 19ms */class Solution {public:    string multiply(string num1, string num2) {        vector<int> ans{0};        vector<int> number1 = convert(num1), number2 = convert(num2);        for (int i = 0; i < number1.size(); ++i) {            vector<int> product = multiply(number1[i], i, number2);            add(ans, product);        }        return convert(ans);    }private:    vector<int> convert(string num) {        vector<int> number;        for (int i = num.size() - 1; i >= 0; --i) {            number.push_back(num[i] - '0');        }        return number;    }    string convert(vector<int> num) {        stringstream ss;        bool zero = true;        for (int i = num.size() - 1; i >= 1; --i) {            if (zero && num[i] == 0) {                continue;            } else {                zero = false;            }            ss << num[i];        }        ss << num[0];        return ss.str();    }    vector<int> multiply(int num1, int zero, vector<int> &num2) {        vector<int> ans;        while (zero--) {            ans.push_back(0);        }        int tmp = 0;        for (int i = 0; i < num2.size(); ++i) {            tmp += num1 * num2[i];            ans.push_back(tmp % 10);            tmp /= 10;        }        if (tmp != 0) {            ans.push_back(tmp);        }        return ans;    }    void add(vector<int> &num1, vector<int> &num2) {        if (num1.size() < num2.size()) {            swap(num1, num2);        }        int tmp = 0;        for (int i = 0; i < num1.size(); ++i) {            tmp += num1[i];            if (i < num2.size()) {                tmp += num2[i];            }            num1[i] = tmp % 10;            tmp /= 10;        }        if (tmp != 0) {            num1.push_back(tmp);        }    }};