[LeetCode] 大数问题,相加和相乘,题 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.
class Solution { public: string multiply(string num1, string num2) { } };
我的解法是每次提取num2的一位,然后和num1相乘,所得结果并入string res中,这样虽然每次都要新定义一个string,效率略低,但是思路比较清晰:把字符串相乘划分成了“字符串和数字相乘” 和 “字符串相加” 两个子问题,分别解决就可以。
处理时先将num1,和num2逆序,使得最低位在[0] 位置,这样处理起来比较方便,不易写错。
class Solution { public: string multiply(string num1, string num2) { if(num1.length() == 0 || num2.length() == 0) return ""; if(num1.length() == 1 && num1[0] == '0') return "0"; if(num2.length() == 1 && num2[0] == '0') return "0"; string res(num1.length() + num2.length(), '0'); int i = 0, j = 0; reverse(num1.begin(), num1.end()); reverse(num2.begin(), num2.end()); for(i = 0; i < num2.length(); ++i){ int offset = i; string tmp(offset + num1.length() + 1, '0'); int add = 0; for(j = 0; j < num1.length(); ++j){ int tmpMul = (num1[j] - '0') * (num2[i] - '0') + add; add = tmpMul / 10; tmp[j + offset] = (tmpMul % 10) + '0'; } tmp[j + offset] = add + '0'; plus(res, tmp); } int countStartZero = 0; //高位'0'的个数 for(i = res.length() - 1; i >= 0 && res[i] == '0'; --i, ++countStartZero); if(countStartZero == res.length()) return "0"; res = res.substr(0, res.length() - countStartZero); reverse(res.begin(), res.end()); return res; } private: void plus(string &s1, string &s2){ int add = 0; for(int i = 0; i < s1.length() && (add > 0 || i < s2.length()); ++i){ int tmp = (s1[i] - '0') + add; if(i < s2.length()) tmp += (s2[i] - '0'); add = tmp / 10; s1[i] = (tmp % 10) + '0'; } } };
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Felix原创,转载请注明出处,感谢博客园!
posted on 2014-05-06 11:06 Felix Fang 阅读(881) 评论(0) 编辑 收藏 举报