Add Binary LeetCode Java

描述
Given two binary strings, return their sum (also a binary string).
For example,

a = "11"

b = "1"

Return ”100”

分析

无

代码

public class AddBinary {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        String str1="111";
        String str2="1";
        System.out.println(addBinary (str1,str2));
}
    public static String addBinary(String str1,String str2) {
        if(str1==""||str2=="") return null;
//        int a1=Integer.parseInt(str1,2);
//        int a2=Integer.parseInt(str2,2);
        
        char[] ch1=str1.toCharArray();
        char[] ch2=str2.toCharArray();
        int a1=0,a2=0;
        for(int i=0;i<ch1.length;i++) {
            a1+=Math.pow(2, i)*(str1.charAt(ch1.length-1-i)-'0');
        }
        
        for(int i=0;i<ch2.length;i++) {
            a2+=Math.pow(2, i)*(str2.charAt(ch2.length-1-i)-'0');
        }
        int num=a1+a2;
        String result=Integer.toBinaryString(num);
        return result;
   }
    
}

 方法二

public static String addBinary(String str1, String str2) {
        if (str1 == "" || str2 == "")
            return null;
int n = Math.max(str1.length(), str2.length());

        char[] res = new char[n];
        int flag = 0;
        int a1 = 0, a2 = 0;
        for (int i = 0; i < n; i++) {
            if (i < str1.length())
                a1 = str1.charAt(str1.length() - 1 - i) - '0';
            else
                a1 = 0;

            if (i < str2.length())
                a2 = str2.charAt(str2.length() - 1 - i) - '0';
            else
                a2 = 0;

            // if ((a1 + a2 + flag)%2==1) {
            // res[n - 1 - i] ='1';
            // flag = (a1 + a2 + flag)/2;
            // }
            // else if((a1 + a2 + flag)%2==0){
            // res[n - 1 - i] = '0';
            // flag = (a1 + a2 + flag)/2;
            // }

            if ((a1 + a2 + flag) == 3) {
                res[n - 1 - i] = '1';
                flag = 1;
            }

            else if (a1 + a2 + flag == 2) {
                res[n - 1 - i] = '0';
                flag = 1;

            } else if (a1 + a2 + flag == 1) {
                res[n - 1 - i] = '1';
                flag = 0;

            } else {
                res[n - 1 - i] = '0';
                flag = 0;
            }
        }

        return (flag == 0) ? new String(res) : "1" + new String(res);
        // return result;
    }