假如str[0]存放的数据类型是字符‘1’，那么减去‘0’就得到一个整形数字1，而不是字符型‘1’，你还是不懂得话，可以查看C程序设计的最后面的“常用字符与ASCLL代码对照表”。字符‘0’的ASCLL值是48，而字符‘1’是49，所以str[0]-'0'相当于49-48=1
public class Solution {
public String addBinary(String a, String b) {
String rea = new StringBuilder(a).reverse().toString();
String reb = new StringBuilder(b).reverse().toString();
StringBuilder result = new StringBuilder();
boolean isCarry= false;
int length = Math.max(rea.length(),reb.length());
for (int i = 0; i <length ; i++) {
int x = i<rea.length()?rea.charAt(i)-'0':0;
int y = i<reb.length()?reb.charAt(i)-'0':0;
int sum= isCarry ? x+y+1: x+y;
isCarry = sum>1 ? true:false;
if(sum%2==0){
result.append("0");
}else {
result.append("1");
}
}
if(isCarry){
result.append("1");
}
return result.reverse().toString();
}
}
Shawngbk
commented
7 years ago
假如str[0]存放的数据类型是字符‘1’，那么减去‘0’就得到一个整形数字1，而不是字符型‘1’，你还是不懂得话，可以查看C程序设计的最后面的“常用字符与ASCLL代码对照表”。字符‘0’的ASCLL值是48，而字符‘1’是49，所以str[0]-'0'相当于49-48=1 public class Solution { public String addBinary(String a, String b) { String rea = new StringBuilder(a).reverse().toString(); String reb = new StringBuilder(b).reverse().toString(); StringBuilder result = new StringBuilder(); boolean isCarry= false; int length = Math.max(rea.length(),reb.length()); for (int i = 0; i <length ; i++) { int x = i<rea.length()?rea.charAt(i)-'0':0; int y = i<reb.length()?reb.charAt(i)-'0':0; int sum= isCarry ? x+y+1: x+y; isCarry = sum>1 ? true:false; if(sum%2==0){ result.append("0"); }else { result.append("1"); } } if(isCarry){ result.append("1"); } return result.reverse().toString(); } }