算法-大整数相乘

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今天在k6k4在线编程做了一道算法题,记录一下:原文:大整数相乘

两个大整数相乘容易溢出,现将整数用字符数组表示,模拟整数相乘的过程。如两个整数字符数组表示为: a = new char[]{'1', '9', '9'}, b = new char[]{'2', '9', '9', '9', '9'}, 结果为:c = new char[]{'5', '9', '6', '9', '8','0','1'}。 注:大整数首位不能为0

输入、输出描述
输入:

两个大整数的字符串表示形式,排列从高位到地位,如321 的数组表示形式:new char[]{'3', '2', '1'}

输出:

两个大整数相乘结果的字符数组表示形式,首位不能为0

Example
输入:

a = new char[]{'1', '9', '9'}
b = new char[]{'2', '9', '9', '9', '9'}

输出:

c = new char[]{'5', '9', '6', '9', '8','0','1'}


解法:


 
  import java.util.*;        public class Main {        public char[] solution(char[] a,char[] b) {    char[] c = new char[a.length + b.length];                int i, j, m, n;            int sum, carry;            m = a.length - 1;            n = b.length - 1;            for (i = m; i >= 0; i--)                a[i] -= '0';            for (i = n; i >= 0; i--)                b[i] -= '0';            carry = 0;            for (i = m + n; i >= 0; i--) {                sum = carry;                if ((j = (i - m)) < 0)                    j = 0;                for (; j <= i && j <= n; j++)                    sum += a[i - j] * b[j];                c[i + 1] = (char) (sum % 10 + '0');                carry = sum / 10;            }            if ((c[0] = (char) (carry + '0')) == '0') {                char[] result = new char[c.length - 1];                for (i = 1; i < c.length; i++) {                    result[i - 1] = c[i];                }                return result;            } else {                return c;            }        }    }

原文:大整数相乘