[LeetCode] 67. Add Binary java

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    /**67. Add Binary      * @param a     * @param b     * @returnString 字符串模拟二进制位相加运算     */       public String addBinary(String a, String b) {            if (a == null || a.length() ==0)                return b;            if (b == null || b.length() == 0)                return a;            String ret = "";            char add = '0';            int i=a.length()-1, j=b.length()-1;            while (i>=0 && j>=0) {                int sum = (a.charAt(i)-'0') + (b.charAt(j)-'0') + (add - '0');                if (sum >= 2) {                    add = '1';                    String s = (sum-2)==0? "0" :"1";                    ret = s+ret;                } else {                    add = '0';                    ret = String.valueOf(sum)+ret;                }                i--;                j--;            }            while(i>=0) {                int sum = (a.charAt(i)-'0') +  (add - '0');                if (sum >= 2) {                    add = '1';                    String s = (sum-2)==0? "0" :"1";                    ret = s+ret;                }else {                    add = '0';                    ret = String.valueOf(sum)+ret;                }                i--;            }            while(j>=0) {                int sum = (b.charAt(j)-'0') +  (add - '0');                if (sum >= 2) {                    add = '1';                    String s = (sum-2)==0? "0" :"1";                    ret = s+ret;                }else {                    add = '0';                    ret = String.valueOf(sum)+ret;                }                j--;            }            if (add != '0') {                ret = String.valueOf(add) + ret;            }            return ret;        }

用add存储每次相加时的进位
记得最后循环结束后，要再次加一次add

改进：

    public String addBinary1(String a, String b) {        int m = a.length();        int n = b.length();        int carry = 0;        String res = "";        int maxLen = Math.max(m, n);        for (int i = 0; i < maxLen; i++) {            int p = 0, q = 0;            if (i < m)                p = a.charAt(m - 1 - i) - '0';            else                p = 0;            if (i < n)                q = b.charAt(n - 1 - i) - '0';            else                q = 0;            int tmp = p + q + carry;            carry = tmp / 2;            res += tmp % 2;        }        return (carry == 0) ? res : "1" + res;    }

二进制加法都是从最低位（从右加到左）。所以对两个字符串要从最后一位开始加，如果遇见长度不一的情况，就把短的字符串高位补0.
每轮计算要加上进位，最后跳出循环后要坚持进位是否为1，以便更新结果。

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