题解 | #二进制求和#

import java.math.BigInteger;
import java.util.*;

/**
 * NC188 二进制求和
 * @author d3y1
 */
public class Solution {
    // 进制
    private final int RADIX = 2;

    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param A string字符串
     * @param B string字符串
     * @return string字符串
     */
    public String binaryAdd (String A, String B) {
        return solution1(A, B);
        // return solution2(A, B);
    }

    /**
     * 模拟法
     * @param A
     * @param B
     * @return
     */
    private String solution1(String A, String B){
        int carry = 0;
        int sum,digitA,digitB,digit;
        StringBuilder result = new StringBuilder();

        int i = A.length()-1;
        int j = B.length()-1;
        while(i>=0 || j>=0 || carry>0){
            digitA = (i>=0)?A.charAt(i--)-'0':0;
            digitB = (j>=0)?B.charAt(j--)-'0':0;
            sum = digitA + digitB + carry;
            digit = sum % RADIX;
            carry = sum / RADIX;
            result.insert(0, digit);
        }

        return result.toString();
    }

    /**
     * 库函数: BigInteger
     * @param A
     * @param B
     * @return
     */
    private String solution2(String A, String B){
        BigInteger bigIntegerA = new BigInteger(A, RADIX);
        BigInteger bigIntegerB = new BigInteger(B, RADIX);

        return bigIntegerA.add(bigIntegerB).toString(RADIX);
    }
}