題目

描述

給定兩個(gè)二進(jìn)制字符串，返回他們的和（用二進(jìn)制表示）。

樣例

a = 11
b = 1
返回 100

解答

思路

1. 字符串拆成字符數(shù)組，遍歷字符數(shù)組進(jìn)行計(jì)算，考慮到兩個(gè)字符串可能長(zhǎng)短不同，先算短的一截（進(jìn)位，兩個(gè)加數(shù)共三個(gè)參數(shù)），再算長(zhǎng)一點(diǎn)的字符串的剩余部分（進(jìn)位，一個(gè)加數(shù)共兩個(gè)參數(shù)）。
2. 抽離按位相加功能，返回參數(shù)中'1'的個(gè)數(shù)count，根據(jù)count進(jìn)行不同處理。
3. 注意字符串，字符數(shù)組處理順序跟數(shù)字相加是反的，計(jì)算時(shí)要從右往左，存儲(chǔ)在緩存字符串中之后也要轉(zhuǎn)置。

代碼

public class Solution {
    /**
     * @param a a number
     * @param b a number
     * @return the result
     */
    public String addBinary(String a, String b) {
        // Write your code here
        StringBuffer sb = new StringBuffer();
        char carry = '0';
        int count;
        char[] as = a.toCharArray();
        char[] bs = b.toCharArray();
        char[] longer = (as.length > bs.length) ? as : bs;
        int minLength = (as.length > bs.length) ? bs.length : as.length;
        for (int i = 0; i < minLength; i++){
            count = add(as[as.length - i - 1], bs[bs.length - i - 1], carry);
            if(count == 0) {carry = '0';sb.append('0');}
            else if(count == 1) {carry = '0';sb.append('1');}
            else if(count == 2) {carry = '1';sb.append('0');}
            else if(count == 3) {carry = '1';sb.append('1');}
            else return "0";
        }
        for(int j = minLength; j < longer.length; j++){
            count = add(longer[longer.length - j - 1], carry, '0');
            if(count == 0) {carry = '0';sb.append('0');}
            else if(count == 1) {carry = '0';sb.append('1');}
            else if(count == 2) {carry = '1';sb.append('0');}
            else return "0";
        }
        if(carry == '1') sb.append('1');
        return sb.reverse().toString();
    }
    public int add(char a, char b, char c){
        short count = 0;
        count += ((a == '0') ? 0 : 1);
        count += ((b == '0') ? 0 : 1);
        count += ((c == '0') ? 0 : 1);
        return count;
    }
}