leetcode -- Add Binary

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

For example,
a = "11"
b = "1"
Return "100".

[解题思路]

将a,b转成10进制相加得到结果再转成二进制，可以过small，跑large时溢出

"10100000100100110110010000010101111011011001101110111111111101000000101111001110001111100001101", "110101001011101110001111100110001010100001101011101010000011011011001011101111001100000011011110011"

"11101000101011001000011011000001100011110011010010011000000000"

"110111101100010011000101110110100000011101000101011001000011011000001100011110011010010011000000000"

public class Solution {
    public String addBinary(String a, String b) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(a == null || a.equals("")){
            return b;
        }
        if(b == null || b.equals("")){
            return a;
        }
        
        long a1 = 0;
        for(int i = 0; i < a.length(); i++){
            a1 = a1 * 2 + (a.charAt(i) - '0');
        }
        long b2 = 0;
        for(int i = 0; i < b.length(); i++){
            b2 = b2 * 2 + (b.charAt(i) - '0');
        }
        
        long sum = a1 + b2;
        if(sum == 0){
            return "0";
        }
        ArrayList<Long> binarySum = new ArrayList<Long>();
        while(sum > 0){
            binarySum.add(sum % 2);
            sum = sum / 2;
        }
        StringBuilder sb = new StringBuilder();
        for(int i = binarySum.size() - 1; i >= 0; i-- ){
            sb.append(binarySum.get(i));
        }
        return sb.toString();
    }
}

字符串操作

1. StringBuilder.reverse()

2. 最后一位的进位需要加上

public class Solution {
    public String addBinary(String a, String b) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(a == null || a.equals("")){
            return b;
        }
        if(b == null || b.equals("")){
            return a;
        }
        StringBuilder a1 = new StringBuilder(a);
        a1.reverse();
        StringBuilder b1 = new StringBuilder(b);
        b1.reverse();
        StringBuilder result = new StringBuilder();
        int len = Math.max(a.length(), b.length());
        int carry = 0;
        for(int i = 0; i < len; i++){
            int t1 = (i >= a.length() ? 0 : (a1.charAt(i) - '0'));
            int t2 = (i >= b.length() ? 0 : (b1.charAt(i) - '0'));
            int t3 = t1 + t2 + carry;
            carry = t3 / 2;
            t3 = t3 % 2;
            result.append(t3);
        }
26         if(carry != 0)
27             result.append(carry);
        result.reverse();
        return result.toString();
    }
}