題目:
Given two binary strings, return their sum (also a binary string).
For example,
a = "11"
b = "1"
Return "100"
.
思路:
兩個二進制位和進位相加的的結果有4種(0,1,2,3),所以只要兩個字符串都逐位從最後一位往前遍歷,並用一個變量保存進位,然後對三者相加結果進行相應處理即可,最後對相加得到的字符串逆序輸出即可,具體細節見代碼吧。
public String addBinary(String a, String b) {
if (a == null || b == null || a.length() == 0 || b.length() == 0) {
return null;
}
int i = a.length() - 1,j = b.length() - 1;
int temp,k = 0;
StringBuilder stringBuilder = new StringBuilder();
//下面和歸併排序的Merge部分有點像
while (i >= 0 && j >= 0) {
temp = a.charAt(i--) - '0' + b.charAt(j--) - '0' + k;
if (temp == 0 || temp == 1) {
stringBuilder.append(temp);
k = 0;
} else {
if (temp == 2) {
stringBuilder.append(0);
k = 1;
} else {
stringBuilder.append(1);
k = 1;
}
}
}
while (i >= 0) {
temp = a.charAt(i--) - '0' + k;
if (temp == 0 || temp == 1) {
stringBuilder.append(temp);
k = 0;
} else {
stringBuilder.append(0);
k = 1;
}
}
while (j >= 0) {
temp = b.charAt(j--) - '0' + k;
if (temp == 0 || temp == 1) {
stringBuilder.append(temp);
k = 0;
} else {
stringBuilder.append(0);
k = 1;
}
}
//如果最後還有進位
if (k == 1) {
stringBuilder.append(1);
}
return stringBuilder.reverse().toString();//逆序輸出
}
再補充個Python寫的簡潔些的版本吧
class Solution(object):
def addBinary(self, a, b):
"""
:type a: str
:type b: str
:rtype: str
"""
i = len(a) - 1
j = len(b) - 1
carry = 0
c = ''
while i >= 0 or j >= 0 or carry > 0:
temp = 0
if i >= 0:
temp += int(a[i])
i -= 1
if j >= 0:
temp += int(b[j])
j -= 1
temp += carry
if temp == 0:
c = '0' + c
carry = 0
elif temp == 1:
c = '1' + c
carry = 0
elif temp == 2:
c = '0' + c
carry = 1
else:
c = '1' + c
carry = 1
return c