【leetcode】AddBinary

解题思路：

• 首先把字符串a和b补齐到同样长度，前面补零
• 然后对应位做加法器，设置一个进位位
• 字符转换成数字需要 -'0'
• 数字转换为字符需要 +'0'
• 结果字符数组需要比a和b多一位
• 最前面的一位为最后的进位位
  • 判断最后的进位如果为1,则设置结果数组第一位为1，创建字符串用result数组全部
  • 如果为0，则设置结果字符数组第一位为0，创建字符串用1到最后
    • 注意
      • 1和0都需要转换为char，不然会乱码，因为是acssi码
      • 创建字符1到最后，是1到result.lenght-1

代码：

package leetcode;

   

/**

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

* example, a = "11" b = "1" Return "100".

*

*/

public class AddBinary {

    public static void main(String[] args) {

        String a = "101";

        String b = "1";

        int max_a_b = Math.max(a.length(), b.length());

        if (a.length() >= b.length()) {

            b = String.format("%0" + max_a_b + "d", Integer.parseInt(b));

        } else {

            a = String.format("%0" + max_a_b + "d", Integer.parseInt(a));

        }

   

        System.out.println("after zeroize string a is " + a + "\n"

                + "after zeroize string b is " + b);

        char[] achar = a.toCharArray();

        char[] bchar = b.toCharArray();

   

        char[] result = new char[Math.max(achar.length, bchar.length) + 1];

        int k = result.length - 1;// k is to the start of result

        int i = max_a_b - 1;

   

        int carry = 0;// 进位

        int aint = 0;

        int bint = 0;

        while (i >= 0) {

            aint = achar[i] - '0';

            bint = bchar[i] - '0';

            if (aint + bint + carry > 1) {

                result[k] = (char) ('0' + aint + bint + carry - 2);

                carry = 1;

   

            } else {

                result[k] = (char) ('0' + aint + bint + carry);

                carry = 0;

   

            }

            k--;

            i--;

        }

        if (carry == 1) {

            result[k] = (char) ((char) carry + '0');

            System.out.println("result is " + new String(result));

        } else {

            result[k] = (char) ((char) carry + '0');

            System.out.println("result is "

                    + new String(result, 1, result.length - 1));

        }

    }

}