1.问题描述

Given two binary strings, return their sum (also a binary string). The input strings are both non-empty and contains only characters 1 or 0. Example 1: Input: a = “11”, b = “1” Output: “100” Example 2: Input: a = “1010”, b = “1011” Output: “10101” 来自 https://leetcode.com/problems/add-binary/description/

2.题目分析

对两个二进制的字符串进行求和运算，需要注意字符和数字的转换，利用字符’0’的ascII码为48进行计算。以及结果是类似于尾插法，因此结果要进行字符串反转。

3.c++代码

//我的代码：(beats 55%) string addBinary(string &a, string &b) { string r; int m = a.length() - 1; int n = b.length() - 1; int c = 0; while (m > -1 || n > -1) { int left = m > -1 ? (int)a[m]-48 : 0;//'0'=48 int right = n > -1 ? (int)b[n]-48 : 0; int sum = left + right + c; c = sum / 2 ; sum = sum %2; r = r + (char)(sum+48); if( m>-1) m--; if (n >-1) n--; } if (c == 1)r = r + '1'; for (int i = 0, j = r.length() - 1; i<j; i++, j--) //字符串反转 { char c = r[i]; r[i] = r[j]; r[j] = c; } return r; } //别人家的代码： string addBinary(string a, string b) { string result = ""; int c = 0; int i = a.size() - 1; int j = b.size() - 1; while (i >= 0 || j >= 0 || c == 1) { c += i >= 0 ? a[i--] - '0' : 0; c += j >= 0 ? b[j--] - '0' : 0; result = char(c % 2 + '0') + result; c /= 2; } return result; }

4.字符串反转

void reverse(char *s, int n) { for (int i = 0, j = n - 1; i < j; i++, j--) { char c = s[i]; s[i] = s[j]; s[j] = c; } }