[LeetCode]第十二题 ：二进制相加

题目描述：

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"

题目解释：

给出两个二进制字符串，返回它们的和（也是二进制字符串）

输入的字符串非空并且只有1和0

题目解法：

1.我的解法：新建一个char数组，用来存放二进制的值；将a和b都转化为char数组；设置一个temp标识进位，从末尾开始遍历，分别去a对应的char、b对应的char和进位char，判断四种情况；遍历结束转化为String型结果进行输出。

class Solution {
    public String addBinary(String a, String b) {
        int length = a.length();
        if(length < b.length()) length = b.length();
        char[] array = new char[length + 1];
        char temp = '0';
        char[] aArray = a.toCharArray();
        char[] bArray = b.toCharArray();
        for(int i = array.length - 1;i>=0;i--) {
            int indexTemp = array.length - 1 - i;
            int aIndex = aArray.length - 1 - indexTemp;
            int bIndex = bArray.length - 1 - indexTemp;
            char aTemp,bTemp;
            if(aIndex >= 0) aTemp = aArray[aIndex];
            else aTemp = '0';
            if(bIndex >= 0) bTemp = bArray[bIndex];
            else bTemp = '0';
            if(temp == '0' && aTemp == '0' && bTemp == '0') {
                array[i] = '0';
                temp = '0';
            } else if(temp == '1' && aTemp == '1' && bTemp == '1') {
                array[i] = '1';
                temp = '1';
            } else if((temp == '1' && aTemp == '0' && bTemp == '0') || (temp == '0' && aTemp == '1' && bTemp == '0') || 
                      (temp == '0' && aTemp == '0' && bTemp == '1')) {
                array[i] = '1';
                temp = '0';
            } else {
                array[i] = '0';
                temp = '1';
            }
        }
        String result = "";
        for(int i = 0;i< array.length;i++){
            if(i == 0) {
                if(array[0] != '0') result += array[i];
            } else{
                result += array[i];
            }
        }
        return result;
    }
}