LeetCode#67. Add Binary

• 题目：两个二进制数相加（String）
• 思路：从低位到高位，两个数对应位相加（考虑进位）
• 难度：Easy
• 代码：

public class Solution {    public String addBinary(String a, String b) {        int len1 = a.length();        int len2 = b.length();        if(len1 == 0){            return b;        }        if(len2 == 0){            return a;        }       int len = len1 > len2 ? len1 : len2;       int[] result = new int[len+1];       int i = len;       int carry = 0;       int sum = 0;       while(len1 > 0 && len2 > 0){           sum = (a.charAt(len1-1) -'0') + (b.charAt(len2-1) - '0') + carry;           carry = sum/2;           result[i] = sum%2;           System.out.print(result[i]);           i--;           len1--;           len2--;       }       while(len1 > 0){           sum = (a.charAt(len1-1) - '0') + carry;           carry = sum/2;           result[i] = sum%2;           i--;           len1--;       }       while(len2 > 0){           sum = (b.charAt(len2-1) - '0') + carry;           carry = sum/2;           result[i] = sum%2;           i--;           len2--;       }       result[i] = carry;       StringBuilder sb = new StringBuilder();       for(int p:result){           if(!(sb.length() == 0 && p == 0)) sb.append(p);       }       return sb.length() == 0? "0" : sb.toString();    }}

自己写的代码太过冗长了，看完Discuss里的简洁代码，觉得自己写的都是什么鬼❓

• Discuss里的代码（不用申明一个数组来存值，直接在一个while循环里就能处理字符串长度不等的问题）

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;//这一行代码好厉害啊            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();    }}