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;
}
}