赵乐ACM

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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.

思路:

大整数运算的思路,模拟乘法运算

 

代码:

class Solution {
public:
    string multiply(string num1, string num2) {
        int len1 = num1.size(), len2 = num2.size(), len = len1 + len2;
        string str(len, '0');
        for (int i = len1 - 1; i >= 0; i--)
        {
            int a = num1[i] - '0';
            for (int j = len2 - 1, k = len2 + i; j >= 0; j--, k--)
            {
                int b = num2[j] - '0';
                int c = str[k] - '0';
                int t = b * a + c;
                str[k] = t % 10 + '0';
                int d = (str[k-1] - '0') + t / 10;
                if (d >= 10)  //开始这里没有等号,检查了很久才发现,细心啊细心
                    str[k-2] = str[k-2] - '0' + d / 10 + '0';
                str[k-1] = d % 10 + '0';
            }
        }
        int x = 0;
        while (str[x] == '0')
            x++;
        if (str.substr(x, len - x) == "")
            return "0";
        return str.substr(x, len - x);

    }
};

 

posted on 2013-08-10 10:45  赵乐ACM  阅读(3044)  评论(1编辑  收藏  举报