题目

代码

public String addStrings(String num1, String num2) {if (num1 == null) {return num2;}if (num2 == null) {return num1;}int min = Math.min(num1.length(), num2.length());int abs = Math.abs(num1.length() - num2.length());// 0.字符串对齐（补0）if (num1.length() == min) {num1 = fillZero(abs) + num1;}if (num2.length() == min) {num2 = fillZero(abs) + num2;}// 1.字符串相加boolean carry = false;boolean lastCarry = false;StringBuilder result = new StringBuilder();for (int i = num1.length() - 1; i >= 0; i--) {int plus = Integer.parseInt(String.valueOf(num1.charAt(i))) + Integer.parseInt(String.valueOf(num2.charAt(i)));if (carry) {plus += 1;}if (plus > 9) {carry = true;} else {carry = false;}result.append(plus % 10);// 并且是最后的进位，需要补1if (i == 0 && carry ) {lastCarry = true;}}// 2.字符串反转String finalString = lastCarry ? "1" + result.reverse() : result.reverse().toString();// 99 + 9System.out.println(String.format("%s + %s = %s", num1, num2, finalString));return finalString;}public String fillZero(int abs) {StringBuilder zero = new StringBuilder();for (int i = 0; i < abs; i++) {zero.append("0");}return zero.toString();}

总结

• 常规的思路就可以解决
• 先把计算做分解
• 高位补0或者截取高位都可以处理，但是细想一下补0更方便