https://leetcode.com/problems/add-strings/c++
Given two non-negative numbers num1 and num2 represented as string, return the sum of num1 and num2.git
Note:函数
The length of both num1 and num2 is < 5100. Both num1 and num2 contains only digits 0-9. Both num1 and num2 does not contain any leading zero. You must not use any built-in BigInteger library or convert the inputs to integer directly.
题目叫“字符串相加”,给定两个非负的由字符串表示整数,求出它们的和。这里须要注意的是,num1和num2字符串的长度都小于5100,这是个很大的数字,确定不能用int来计算。数字只包含0-9,也就是说没有负号,没有小数点等,彻底当作整形来计算。不容许使用内建的大整形的库函数。ui
实现一:思路比较简单,既然不能把字符串转为整形来计算,那就一个字符一个字符取出来相加,算好进位便可。翻译
public String addStrings(String num1, String num2) { String bigStr = null; String smallStr = null; if (num1.length() >= num2.length()) { bigStr = num1; smallStr = num2; } else { bigStr = num2; smallStr = num1; } int big = bigStr.length(); int small = smallStr.length(); int carry = 0; char []ra = new char[big + 1]; for (int i = 0; i < small; i++) { int b = Character.getNumericValue(bigStr.charAt(big - i - 1)); int s = Character.getNumericValue(smallStr.charAt(small - i - 1)); ra[big - i] = Character.forDigit((b + s + carry) % 10, 10); carry = (b + s + carry) / 10; } for (int i = 0; i < big - small; i++) { int b = Character.getNumericValue(bigStr.charAt(big - small - i - 1)); ra[big - small - i] = Character.forDigit((b+ carry) % 10, 10); carry = (b + carry)/ 10; } if (carry != 0) { ra[0] = Character.forDigit(carry % 10, 10); } String ret = new String(ra).trim(); return ret; }
实现二:从disguss选出来的最简解,我把它从c++翻译成了Java。与实现一并无什么本质不一样,只是代码更简洁,更清晰。牛人太多,佩服。code
public String addStrings(String num1, String num2) { int i = num1.length() - 1, j = num2.length() - 1, carry = 0; String res = ""; while (i >= 0 || j >= 0) { if (i >= 0) carry += num1.charAt(i--) - '0'; if (j >= 0) carry += num2.charAt(j--) - '0'; res = Integer.toString(carry % 10) + res; carry /= 10; } return carry != 0 ? "1" + res : res; }