leetcode解题方案--043-- Multiply Strings

题目

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

Note:

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.

分析

两个可能狠狠狠大的数相乘。考察乘法的分解。如何分解乘法再把结果想加？

我用了递归，直到把乘法分解成一位数成一位数为止。

public static String multiply(String num1, String num2) {
char[] n1 = num1.toCharArray();
char[] n2 = num2.toCharArray();
if (n1.length == 1 && n2.length == 1) {
return String.valueOf((n1[0] - '0') * (n2[0] - '0'));
}
else if (n2.length == 1) {
long sum = 0;
for (int i = 0; i < n1.length; i++) {
sum = sum*10;
sum = sum + Integer.parseInt(multiply(num2,String.valueOf(num1.substring(i, i+1))));
}
return String.valueOf(sum);
}
else if (n1.length == 1) {
return multiply(num2, num1);
} else {
String sum = "";
for (int i = 0; i < n1.length; i++) {
sum = sum+"0";
String xx = multiply(num2,String.valueOf(num1.substring(i, i+1)));
char[] big = (sum.length()>xx.length()?sum:xx).toCharArray();
char[] small = (sum.length()<=xx.length()?sum:xx).toCharArray();
int add = 0;
for (int k = 0;k<small.length;k++) {
int a = small[small.length-1-k]-'0';
int b = big[big.length-1-k]-'0';
int addd = a+b+add;
big[big.length-1-k] = (char) (addd%10+'0');

4000
add = addd/10;
}
int p = big.length - small.length -1;
while (add!=0 && p>=0) {
big[p] = (char) ((big[p]+add-'0')%10+'0');
add=big[p]+add-'0';
add = add/10;
p--;
}
sum = new String(big);
if (add!=0) {
sum ="1"+sum;
}
}
return String.valueOf(sum);
}
}

迄今以来惟一一次晒了ac details，想哭。。

简单分析了一下原因，主要还是拆箱装箱和string的拼接。

比较好的地方是中间运算用了数组，

比较渣的地方是 为了不想多写一个方法，全程string。

建议用stringbuffer来递归，只是需要重写一个递归函数。