二进制求和

给定两个二进制字符串，返回他们的和（用二进制表示）。

输入为非空字符串且只包含数字 1 和 0。

示例 1:

输入: a = "11", b = "1"
输出: "100"
示例 2:

输入: a = "1010", b = "1011"
输出: "10101"

转为数字的方法，数值特别大的情况会有溢出

class Solution {
    public String addBinary(String a, String b) {
        int len1 = a.length();
        int len2 = b.length();
        double sum1 = 0;
        double sum2 = 0;
        
        double count = 0;
        for (int i = len1 - 1; i >= 0; i--) {
            if (a.charAt(i) == '1') {
                double temp = Math.pow(2, count);
                sum1 += temp;
            } 
            count++;
        }
        System.out.println(sum1);
        count = 0;
        for (int j = len2 - 1; j >= 0; j--) {
            if (b.charAt(j) == '1') {
                double temp = Math.pow(2, count);
                sum2 += temp;
            } 
            count++;
        }
        System.out.println(sum2);

        double sum = 0;
        sum = sum1 + sum2;
        if (sum == 0) {
            return "0";
        }
        StringBuffer result = new StringBuffer("");
        while (sum != 1) {
            if (sum % 2 == 0) {
                result.append("0");
            } else {
                result.append("1");

            }
            sum = (int)sum / 2;
        }

        result.append("1");

        return result.reverse().toString();
    }
}

class Solution {
    public String addBinary(String a, String b) {
        //0 + 1 = 1; 0 + 0 = 0; 1 + 1 = 10(进位)
        int lenA = a.length();
        int lenB = b.length();
        int i = lenA - 1 , j = lenB - 1;
        String result = "";
        int carry = 0;//进位
        while (i >= 0 || j >= 0) {
            int tempA, tempB;
            if (i >= 0) {
                tempA = a.charAt(i) - '0';
                i--;
            } else {
                tempA = 0;
            }

            if (j >= 0) {
                tempB = b.charAt(j) - '0';
                j--;
            } else {
                tempB = 0;
            }

            int sum = tempA + tempB + carry;
            result = sum % 2 + result;
            carry = sum / 2;
        }
        if (carry > 0) {
            result = '1' + result;
        }
        return result;
    }
}

将String换为StringBuilder，速度明显提升

class Solution {
    public String addBinary(String a, String b) {
        //0 + 1 = 1; 0 + 0 = 0; 1 + 1 = 10(进位)
        int lenA = a.length();
        int lenB = b.length();
        int i = lenA - 1 , j = lenB - 1;
        StringBuilder result = new StringBuilder("");
        int carry = 0;//进位
        while (i >= 0 || j >= 0) {
            int tempA, tempB;
            if (i >= 0) {
                tempA = a.charAt(i) - '0';
                i--;
            } else {
                tempA = 0;
            }

            if (j >= 0) {
                tempB = b.charAt(j) - '0';
                j--;
            } else {
                tempB = 0;
            }

            int sum = tempA + tempB + carry;
            result.append(sum % 2);
            carry = sum / 2;
        }
        if (carry > 0) {
            result.append('1');
        }
        return result.reverse().toString();
    }
}