标签：library ever rect blog ring bsp input 反转 没有想到
Given two non-negative numbers num1 and num2 represented as string, return the sum of num1 and num2.
Note:
num1 and num2 is < 5100.num1 and num2 contains only digits 0-9.num1 and num2 does not contain any leading zero.
我的第一个想法就是选好两个字符串，然后模拟加法进位，然后用一个字符串接着，同时记住进位，最后返回这个字符串的反转即可。
没有想到什么比较好的解决思路
public class Solution { public String addStrings(String num1, String num2) { StringBuilder sb = new StringBuilder(); int carry = 0; for(int i = num1.length() - 1, j = num2.length() - 1; i >= 0 || j >= 0; i--, j--){ int x = i < 0 ? 0 : num1.charAt(i) - ‘0‘; int y = j < 0 ? 0 : num2.charAt(j) - ‘0‘; sb.append((x + y + carry) % 10); carry = (x + y + carry) / 10; } if(carry != 0) sb.append(carry); return sb.reverse().toString(); } }
标签：library ever rect blog ring bsp input 反转 没有想到
原文地址：http://www.cnblogs.com/linkstar/p/5998891.html
