[LeetCode] 67. Add Binary 二进制数相加
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"
解法: 从最低位加到最高位,当前位相加结果是%2,进位是/2,记得处理每一次的进位和最后一次的进位,最后反向输出字符。
Java:
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;
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();
}
}
Java:
public class AddBinary {
public String addBinary(String a, String b) {
StringBuilder result = new StringBuilder();
int pointerA = a.length()-1;
int pointerB = b.length()-1;
int carry = 0;
while(pointerA>=0 || pointerB>=0){
int sum = carry;
if(pointerA>=0){
sum += (a.charAt(pointerA)-'0');
pointerA--;
}
if(pointerB>=0){
sum += (b.charAt(pointerB)-'0');
pointerB--;
}
result.append(sum%2);
carry = sum/2;
}
if(carry!=0){
result.append('1');
}
return result.reverse().toString();
}
}
Python:
class Solution:
# @param a, a string
# @param b, a string
# @return a string
def addBinary(self, a, b):
result, carry, val = "", 0, 0
for i in xrange(max(len(a), len(b))):
val = carry
if i < len(a):
val += int(a[-(i + 1)])
if i < len(b):
val += int(b[-(i + 1)])
carry, val = val / 2, val % 2
result += str(val)
if carry:
result += str(carry)
return result[::-1]
Python:
class Solution:
def addBinary(self, a, b):
if len(a)==0: return b
if len(b)==0: return a
if a[-1] == '1' and b[-1] == '1':
return self.addBinary(self.addBinary(a[0:-1],b[0:-1]),'1')+'0'
if a[-1] == '0' and b[-1] == '0':
return self.addBinary(a[0:-1],b[0:-1])+'0'
else:
return self.addBinary(a[0:-1],b[0:-1])+'1'
Python:
class Solution(object):
def addBinary(self, a, b):
"""
:type a: str
:type b: str
:rtype: str
"""
res = ''
carry = 0
i, j = len(a) - 1, len(b) - 1
while i >= 0 or j >= 0:
su = carry
if i >= 0:
su += int(a[i])
if j >= 0:
su += int(b[j])
carry = su / 2
res += str(su % 2)
i -= 1
j -= 1 if carry > 0:
res += str(carry) return res[::-1]
C++ 1:
class Solution {
public:
string addBinary(string a, string b) {
string res;
size_t res_len = max(a.length(), b.length()) ; size_t carry = 0;
for (int i = 0; i < res_len; ++i) {
const size_t a_bit_i = i < a.length() ? a[a.length() - 1 - i] - '0' : 0;
const size_t b_bit_i = i < b.length() ? b[b.length() - 1 - i] - '0' : 0;
size_t sum = carry + a_bit_i + b_bit_i;
carry = sum / 2;
sum %= 2;
res.push_back('0' + sum);
}
if (carry) {
res.push_back('0' + carry);
}
reverse(res.begin(), res.end()); return res;
}
};
C++ 2:
class Solution {
public:
string addBinary(string a, string b) {
string res = "";
int m = a.size() - 1, n = b.size() - 1, carry = 0;
while (m >= 0 || n >= 0) {
int p = m >= 0 ? a[m--] - '0' : 0;
int q = n >= 0 ? b[n--] - '0' : 0;
int sum = p + q + carry;
res = to_string(sum % 2) + res;
carry = sum / 2;
}
return carry == 1 ? "1" + res : res;
}
};
Followup:
如果不是二进制相加,而是十六进制相加呢?只要把算法中的除2和余2换成16,并添加相应的十六进制字母就行了。
如果是带小数的?
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