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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)); }}
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