LintCode：M-大数相乘

LintCode链接

Given two non-negative integers num1 and num2represented as strings, return the product of num1 and num2

您在真实的面试中是否遇到过这个题？ 

Yes

样例

• The length of both num1 and num2 is < 110.
• Both num1 and num2 contains only digits 0-9.
• Both num1 and num2 does not contain any leading zero.
• You must not use any built-in BigInteger library or convert the inputs to integer directly.

public class Solution {    /**     * @param num1 a non-negative integers     * @param num2 a non-negative integers     * @return return product of num1 and num2     */    public String multiply(String num1, String num2) {                //保证num1长度>num2        if(num1.length()<num2.length()){            return multiply(num2,num1);        }        int n2 = num2.length();        char[] nums2 = (new StringBuffer(num2)).reverse().toString().toCharArray();        String s1 = singleMul(num1, Integer.valueOf(nums2[0]+""),0);        for(int i=1; i<n2; i++){            String s2 = singleMul(num1, Integer.valueOf(nums2[i]+""),i);            s1 = sum(s1,s2);        }                return s1;    }    String sum(String num1, String num2){        StringBuffer bf = new StringBuffer("");        char[] num1s = (new StringBuffer(num1)).reverse().toString().toCharArray();        char[] num2s = (new StringBuffer(num2)).reverse().toString().toCharArray();        int up=0;        int i=0;        while(i<num1.length() && i<num2.length()){            int sum = Integer.valueOf(num1s[i]+"")+Integer.valueOf(num2s[i]+"")+up;            up = sum/10;            bf.append(sum%10+"");            i++;        }        while(i<num1.length()){            int sum = Integer.valueOf(num1s[i]+"")+up;            up = sum/10;            bf.append(sum%10+"");            i++;        }        while(i<num2.length()){            int sum = Integer.valueOf(num2s[i]+"")+up;            up = sum/10;            bf.append(sum%10+"");            i++;        }        if(up>0)            bf.append(up+"");                return bf.reverse().toString();    }        //第k位上的数knum与num相乘的结果    String singleMul(String num, int knum, int k){        if(knum==0)            return "0";        StringBuffer bf = new StringBuffer("");        char[] chars = new StringBuffer(num).reverse().toString().toCharArray();        int up=0;        for(int i=0; i<chars.length; i++){            int mul = (Integer.valueOf(chars[i]+""))*knum+up;            up = mul/10;            bf.append(mul%10+"");        }        String res = (up>0?up+"":"")+bf.reverse().toString()+getZeros(k);        return res;    }        String getZeros(int k){        StringBuffer bf = new StringBuffer("");        while(k>0){            bf.append("0");            k--;        }        return bf.toString();    }}