*LeetCode--Add Binary

Add Binary

二进制加法  注意考虑本位进位

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"

 

自己的方法:

就是将加法分成进位和本位,然后递归的调用加法,然后当进位部分为 0 时,则为结果。

但是很麻烦

class Solution {
    public String addBinary(String a, String b) {
        if(a == null || a.length() == 0 || b == null || b.length() == 0) return "";
        String str = help(a, b);
        int index = str.indexOf("1");
        if(index == -1) return "0";
        return str.substring(index);
    }

    public static String help(String a, String b){
        if(b.indexOf("1") == -1) return a;
        StringBuilder sum = new StringBuilder();
        StringBuilder carry = new StringBuilder();

        if(a.length() != b.length()){
            String max = a.length() > b.length() ? a : b;
            String min = a.length() < b.length() ? a : b;
            a = max;
            char[] ch = new char[max.length() - min.length()];
            Arrays.fill( ch, '0' );
            b = String.valueOf(ch) + min;
        }
        
        for(int i = 0; i < a.length() && i < b.length(); i++){
      
            if(a.charAt(i) == '1' && b.charAt(i) == '1'){
                carry.append("1");
                sum.append("0");
        
            } else if(a.charAt(i) == '0' && b.charAt(i) == '0'){
                carry.append("0");
                sum.append("0");
           
            } else{
                carry.append("0");
                sum.append("1");
           
            }
        }

        return help(sum.toString(), carry.append("0").toString());
    }
}

  

在discuss区看到的一个简单的方法:

public class Solution {
    public String addBinary(String a, String b) {
        StringBuilder sb = new StringBuilder();
        int i = a.length() - 1, j = b.length() -1, carry = 0;
        while (i >= 0 || j >= 0) {
            int sum = carry;
            if (j >= 0) sum += b.charAt(j--) - '0';
            if (i >= 0) sum += a.charAt(i--) - '0';
            sb.append(sum % 2);
            carry = sum / 2;
        }
        if (carry != 0) sb.append(carry);
        return sb.reverse().toString();
    }
}

  

 

posted @ 2018-05-24 16:28  SkyeAngel  阅读(207)  评论(0编辑  收藏  举报