Problem

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

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

題目的意思就是按二進位制的方式計算兩個數相加。

Java 實現

package com.coderli.leetcode.algorithms.easy;

/**
 * Given two binary strings, return their sum (also a binary string).
 * <p>
 * For example,
 * a = "11"
 * b = "1"
 * Return "100".
 *
 * @author OneCoder 2017-11-07 21:26
 */
 

public class AddBinary {

    public static void main(String[] args) {
        AddBinary addBinary = new AddBinary();
        System.out.println(addBinary.addBinary("11", "1"));
        System.out.println(addBinary.addBinary("11", "11"));
        System.out.println(addBinary.addBinary("11", "111"));
 

        System.out.println(addBinary.addBinary("1", "111"));
        System.out.println(addBinary.addBinary("0", "0"));
    }

    public String addBinary(String a, String b) {
        int aLength = a.length();
        int bLength = b.length();
        int length = aLength >= bLength ? aLength : bLength
 
;
        char[] chars = new char[length];
        int carryDigit = 0;
        for (int i = 0; i < length; i++) {
            int tempSum = 0;
            if (i < aLength && i < bLength) {
                tempSum = a.charAt(aLength - 1 - i) + b.charAt(bLength - 1 - i) + carryDigit - 96;
            } else if (aLength > bLength) {
                tempSum = a.charAt(aLength - 1 - i) + carryDigit - 48;
            } else if (aLength < bLength) {
                tempSum = b.charAt(bLength - 1 - i) + carryDigit - 48;
            }
            carryDigit = tempSum >= 2 ? 1 : 0;
            chars[length - i - 1] = tempSum % 2 == 0 ? (char) 48 : (char) 49;
        }
        String result = new String(chars);
        if (carryDigit != 0) {
            result = "1" + result;
        }
        return result;
    }

}

分析

跟題目Plus One很相近，只是把固定的加1變成了另一個字串。處理方式都完全一致。只是額外需要注意的是char轉int 利用 ‘0’ = 48 額外計算一下即可。