[LeetCode]--67. Add Binary

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

For example,
a = “11”
b = “1”
Return “100”.

二进制算法，加上对字符串的处理。

public String addBinary(String a, String b) {
        if ((a.length() == 0 || a.isEmpty())
                && (b.length() == 0 || b.isEmpty()))
            return null;
        if (a.length() == 0 || a.isEmpty())
            return b;
        if (b.length() == 0 || b.isEmpty())
            return a;
        int i = a.length() - 1, j = b.length() - 1, sum, m = 0;
        int[] temp = new int[Math.max(a.length(), b.length()) + 1];
        int k = temp.length;
        while (i >= 0 && j >= 0) {
            sum = a.charAt(i) - '0' + b.charAt(j) - '0' + m;
            temp[--k] = sum % 2;
            m = sum / 2;
            i--;
            j--;
        }
        while (i >= 0) {
            sum = a.charAt(i) - '0' + m;
            temp[--k] = sum % 2;
            m = sum / 2;
            i--;
        }
        while (j >= 0) {
            sum = b.charAt(j) - '0' + m;
            temp[--k] = sum % 2;
            m = sum / 2;
            j--;
        }
        if (m == 1)
            temp[0] = 1;
        String str = "";
        for (int l = 0; l < temp.length; l++) {
            if (l == 0 && temp[l] == 0)
                continue;
            str = str.concat(Integer.toBinaryString(temp[l]));
        }
        return str;
    }

看了一下专业人士写的，跟我差不太多，还是有一点可以借鉴。如果没必要保持原有的字符串啥的，我们可以对不知道大小的情况先进行排序，就知道哪个遍历先完了。而且字符串实现了+号的重载，没必要用concat方法的其实，直接用加号就行。

public String addBinary(String a, String b) {
        if(a.length() < b.length()){
            String tmp = a;
            a = b;
            b = tmp;
        }

        int pa = a.length()-1;
        int pb = b.length()-1;
        int carries = 0;
        String rst = "";

        while(pb >= 0){
            int sum = (int)(a.charAt(pa) - '0') + (int)(b.charAt(pb) - '0') + carries;
            rst = String.valueOf(sum % 2) + rst;
            carries = sum / 2;
            pa --;
            pb --;
        }

        while(pa >= 0){
            int sum = (int)(a.charAt(pa) - '0') + carries;
            rst = String.valueOf(sum % 2) + rst;
            carries = sum / 2;
            pa --;
        }       

        if (carries == 1)
            rst = "1" + rst;
        return rst;
    }