[leetcode]Multiply Strings

高精度大数相乘。首先是用数组存大数，然后关键的是ans[i+j] += num1[i]*num2[j]。数组中的数这时可能会大于10，就扫一遍做进位处理。

编写中犯过的两个错：1.ans[i+j] += num1[i]*num2[j]没有写+；2.略过末尾的0时，忘记判断index>=0；3.如果结果是0，就是所有的都是0时，要单独返回“0”

参考：http://blog.csdn.net/u011328276/article/details/9464817

public class Solution {
    public String multiply(String num1, String num2) {
        int ans[] = new int[num1.length()+num2.length()];
        int len1 = num1.length();
        int len2 = num2.length();
        for (int i = 0; i < len1; i++)
        {
            for (int j = 0; j < len2; j++)
            {
                ans[i+j] += (num1.charAt(len1-i-1) - '0') * 
                            (num2.charAt(len2-j-1) - '0');
            }
        }
        for (int i = 0; i < len1+len2; i++)
        {
            if (ans[i] >= 10)
            {
                ans[i+1] += ans[i] / 10;
                ans[i] = ans[i] % 10;
            }
        }
        StringBuffer sb = new StringBuffer();
        int i = len1+len2-1;
        while (i >= 0 && ans[i] == 0)
        {
            i--;
        }
        for (; i >= 0; i--)
        {
            sb.append(ans[i]);
        }
        String result = sb.toString();
        if (result.length() == 0) return "0";
        else return result;
    }
}

second time:

class Solution {
public:
    string multiply(string num1, string num2) {
        reverse(num1.begin(), num1.end());
        reverse(num2.begin(), num2.end());
        vector<char> ans(num1.size() + num2.size());
        for (int i = 0; i < num1.size(); i++)
        {
            for (int j = 0; j < num2.size(); j++)
            {
                ans[i+j] += (num1[i] - '0') * (num2[j] - '0');
                ans[i+j+1] += ans[i+j] / 10;
                ans[i+j] = ans[i+j] % 10;
            }
        }
        string s;
        bool nonzero = false;
        for (int i = ans.size()-1; i >= 0; i--)
        {
            if (!nonzero && ans[i] == 0)
                continue;
            nonzero = true;
            s.push_back((char) (ans[i]+'0'));
        }
        if (s == "") return "0";
        return s;
    }
};