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:
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Input: a = "11", b = "1"
Output: "100"
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Example 2:
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Input: a = "1010", b = "1011"
Output: "10101"
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解法
与模拟加法实现的过程相同，只不过是二进制而已，注意不同长度的字符相加和进位的处理，同时记得处理多余的前导0。
具体代码如下：
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class {
public String addBinary(String a, String b) {
StringBuffer res = new StringBuffer();
StringBuffer x = new StringBuffer(a);
x.reverse();
StringBuffer y = new StringBuffer(b);
y.reverse();
if (x.length() < y.length()) {
int add0 = y.length() - x.length();
while (add0 > 0) {
x.append('0');
add0--;
}
} else {
int add0 = x.length() - y.length();
while (add0 > 0) {
y.append('0');
add0--;
}
}
int tmp = 0;
for (int i = 0; i < x.length(); i++) {
if (x.charAt(i) == '0' && y.charAt(i) == '1') {
if (tmp == 0) {
res.append('1');
} else {
res.append('0');
tmp = 1;
}
}
if (x.charAt(i) == '1' && y.charAt(i) == '0') {
if (tmp == 0) {
res.append('1');
} else {
res.append('0');
tmp = 1;
}
}
if (x.charAt(i) == '0' && y.charAt(i) == '0') {
if (tmp == 0) {
res.append('0');
} else {
res.append('1');
tmp = 0;
}
}
if (x.charAt(i) == '1' && y.charAt(i) == '1') {
if (tmp == 0) {
res.append('0');
tmp = 1;
} else {
res.append('1');
tmp = 1;
}
}
}
if (tmp == 1) {
res.append('1');
}
res.reverse();
while (res.charAt(0) == '0' && res.length() > 1) {
res.deleteCharAt(0);
}
return res.toString();
}
}
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