Leetcode 67. Add Binary -- 给定两个01字符串，求和，输出结果字符串

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

The input strings are both non-empty and contains only characters 1 or 0.

Example 1:

Input: a = "11", b = "1"
Output: "100"

Example 2:

Input: a = "1010", b = "1011"
Output: "10101"

方法1：

/**
 * <p>Title: </p>
 * <p>Description: </p>
 *
 * @CreateTime 2019/1/10 22:01
 */
public class Leetcode_67_AddBinary {

    public static void main(String[] args) {

//        Integer a = Integer.parseInt("11", 2);//3
//        System.out.println(Integer.toBinaryString(a));//11

        Leetcode_67_AddBinary leetcode_67_addBinary = new Leetcode_67_AddBinary();
//        String result = leetcode_67_addBinary.addBinary("111001", "1");//111010
        String result = leetcode_67_addBinary.addBinary("111", "1");//1000
        System.out.println(result);
    }


    /**
     * 3 ms, faster than 92.93%
     * @param a
     * @param b
     * @return
     */
    public String addBinary(String a, String b) {
        int inc = 0;
        int aLength = a.length();
        int bLength = b.length();
        char aC;
        char bC;
        int maxLength = Math.max(aLength, bLength);
        int[] values = new int[maxLength + 1];//最大只会比最大值的位数大1位
        for (int i = 0; i < maxLength; i++) {
            if (i < aLength) {
                aC = a.charAt(aLength - 1 - i);
            } else {
                aC = '0';
            }

            if (i < bLength) {
                bC = b.charAt(bLength - 1 - i);
            } else {
                bC = '0';
            }

            if (aC == '1' && bC == '1') {
                values[i] = 0 + inc;
                inc = 1;
            }

            if ((aC == '0' && bC == '1') || (aC == '1' && bC == '0')) {
                values[i] = 1 + inc;
                if (values[i] == 2) {
                    values[i] = 0;
                    inc = 1;
                } else {
                    values[i] = 1;
                    inc = 0;
                }
            }

            if (aC == '0' && bC == '0') {
                values[i] = 0 + inc;
                inc = 0;
            }
        }//for

        if (inc == 1) {//如果计算完所有位时仍有进位，则最高位需要设置为1
            values[maxLength] = 1;
        } else {//如果计算完所有位时没有进位，则最高位是0，maxLength自动减1，遍历时不再输出该位
            maxLength = maxLength - 1;//说明最高位是0
        }

        StringBuilder sb = new StringBuilder();
        for (int i = maxLength; i >= 0; i--) {
            sb.append(values[i]);
        }
        return sb.toString();
    }


    /**
     * 如果  a  b 是很长的 01 字符串，会直接报错
     *
     * @param a
     * @param b
     * @return
     */
    public String addBinary2(String a, String b) {
        Long c = Long.parseLong(a, 2);
        Long d = Long.parseLong(b, 2);
        String s = Long.toBinaryString(c + d);
        return s;
    }


    /**
     * 如果  a  b 是很长的 01 字符串，会直接报错
     *
     * @param a
     * @param b
     * @return
     */
    public String addBinary3(String a, String b) {
        Integer c = Integer.parseInt(a, 2);
        Integer d = Integer.parseInt(b, 2);
        String s = Integer.toBinaryString(c + d);
        return s;
    }

}

方法2：

待续