标签：number 问题 str res height nbsp png rar ace
2018-09-27 22:42:13
一、大整数加法
问题描述：
问题求解：
public String addStrings2(String num1, String num2) {
if (num1.length() < num2.length()) return addStrings(num2, num1);
char[] n1 = num1.toCharArray(), n2 = num2.toCharArray();
int m = n1.length, n = n2.length;
int carry = 0, i = 1, sum = 0;
while (i <= n) {
sum = n1[m - i] + n2[n - i] - ‘0‘ * 2 + carry;
carry = sum / 10;
n1[m - i] = (char) (sum % 10 + ‘0‘);
i++;
}
while (carry > 0 && i <= m) {
sum = n1[m - i] + carry - ‘0‘;
carry = sum / 10;
n1[m - i] = (char) (sum % 10 + ‘0‘);
i++;
}
return carry == 0 ? new String(n1) : "1" + new String(n1);
}
二、大整数乘法
问题描述：
问题求解：
`num1[i] * num2[j]` will be placed at indices `[i + j`, `i + j + 1]`
public String multiply(String num1, String num2) {
int len1 = num1.length();
int len2 = num2.length();
int[] res = new int[len1 + len2];
for (int i = len1 - 1; i >= 0; i--) {
for (int j = len2 - 1; j >= 0; j--) {
int mul = (num1.charAt(i) - ‘0‘) * (num2.charAt(j) - ‘0‘);
int p1 = i + j;
int p2 = i + j + 1;
int sum = mul + res[p2];
res[p1] += sum / 10;
res[p2] = sum % 10;
}
}
StringBuffer sb = new StringBuffer();
for (int i = 0; i < res.length; i++) if (!(sb.length() == 0 && res[i] == 0)) sb.append(res[i]);
return sb.length() == 0 ? "0" : sb.toString();
}
标签：number 问题 str res height nbsp png rar ace
原文地址：https://www.cnblogs.com/TIMHY/p/9716169.html
