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LeetCode 043 Multiply Strings

时间:2015-02-07 21:40:04      阅读:162      评论:0      收藏:0      [点我收藏+]

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题目要求: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.

 

分析:

参考网址:http://blog.csdn.net/pickless/article/details/9235907

利用竖式的思想,进行乘法运算。

 

        3    4

*      1    3

——————

       9      12

3     4

——————

3     13     12

 

再处理进位:

3 13 12 - > 3 14 2 -> 4 4 2

 

代码如下:

class Solution {
public:
    string multiply(string num1, string num2) {

        int flag = 1;
        
        //先处理符号
        if (num1[0] == - || num2[0] == -) {
            if (num1[0] == -) {
                flag *= -1;
                num1 = num1.substr(1, num1.size() - 1);
            }
            if (num2[0] == -) {
                flag *= -1;
                num2 = num2.substr(1, num2.size() - 1);
            }
        }
        
        int length = num1.size() + num2.size() + 1;
        int s[length];
        
        memset(s, 0, sizeof(int) * length);
        
        int i = 0, temp = 0;
        
        for (i = 0; i < num1.size(); i++) {
            for (int j = 0; j < num2.size(); j++) {
                s[(num1.size() - i - 1) + (num2.size() - j - 1)] += (num1[i] - 0) * (num2[j] - 0);
            }
        }
        
        //进位
        for (i = 0; i < length; i++) {
            s[i + 1] += s[i] / 10;
            s[i] = s[i] % 10;
        }
        
        for (i = 0; i < length / 2; i++) {
            temp = s[i];
            s[i] = s[length - i - 1];
            s[length - i - 1] = temp;
        }
        
        string ans = flag < 0 ? "-" : "";
        for (i = 0, temp = -1; i < length; i++) {
            if (s[i] != 0) {
                temp = 1;
            }
            if (temp > 0) {
                ans += s[i] + 0;
            }
        }
        return ans == "" ? "0" : ans;
    }
};

 

LeetCode 043 Multiply Strings

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原文地址:http://www.cnblogs.com/510602159-Yano/p/4279270.html

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