大整数求和,思路:数组存储即可 步骤: 1、创建两个整数数组,数组长度为最大长度整数位数+1。然后分别将两个大整数的值存储到数组中,整数的个位存与数组下标0的位置,最高位位与数组的尾部,方便从左到右的访问数组的习惯 2、创建结果数组,长度是较大数的位数+1,+1是给最高位进位预留的 3、遍历两个数组,从左到右按照对应下标把元素两两相加,就像小学生计算竖式一样 4、把result数组的全部元素再次逆序,去掉首位的0,就是最终结果 时间复杂度:每个步骤的时间复杂度都是O(n)所以整体时间复杂度是O(n)
public class BigNumbersSum {
public static String bigNumbersSum(String bigNumberA, String bigNumberB) {
//1、创建两个数组,并赋值
int maxLength = bigNumberA.length() > bigNumberB.length() ? bigNumberA.length() : bigNumberB.length();
int[] arrayA = new int[maxLength + 1];
int[] arrayB = new int[maxLength + 1];
for (int i = 0; i < bigNumberA.length(); i++) {
arrayA[i] = bigNumberA.charAt(bigNumberA.length() - 1 - i) - '0';
}
for (int i = 0; i < bigNumberB.length(); i++) {
arrayB[i] = bigNumberB.charAt(bigNumberB.length() - 1 - i) - '0';
}
//2、构建结果数组 result
int[] result = new int[maxLength + 1];
//3、遍历两个数组,从左到右相加
for (int i = 0; i < maxLength; i++) {
int temp = result[i];
temp += arrayA[i];
temp += arrayB[i];
if (temp >= 10) {
temp = temp - 10;
result[i + 1] = 1;
}
result[i] = temp;
}
//4、反转int,去掉前面的0
boolean findFirst = false;
StringBuilder stringBuilder = new StringBuilder(maxLength);
for (int i = result.length - 1; i >= 0; i--) {
if (!findFirst) {
if (result[i] == 0) {
continue;
}
findFirst = true;
}
stringBuilder.append(result[i]);
}
return stringBuilder.toString();
}
public static void main(String[] args) {
String bigNumberA = "20000000020000";
String bigNumberB = "80000000080000";
System.out.println(bigNumbersSum(bigNumberA, bigNumberB));
}
}