给你两个二进制字符串 a
和 b
,以二进制字符串的形式返回它们的和。
示例 1:
输入:a = "11", b = "1" 输出:"100"
示例 2:
输入:a = "1010", b = "1011" 输出:"10101"
提示:
1 <= a.length, b.length <= 104
a
和b
仅由字符'0'
或'1'
组成- 字符串如果不是
"0"
,就不含前导零
class Solution:
def addBinary(self, a: str, b: str) -> str:
return bin(int(a, 2) + int(b, 2))[2:]
class Solution:
def addBinary(self, a: str, b: str) -> str:
ans = []
i, j, carry = len(a) - 1, len(b) - 1, 0
while i >= 0 or j >= 0 or carry:
carry += (0 if i < 0 else int(a[i])) + (0 if j < 0 else int(b[j]))
carry, v = divmod(carry, 2)
ans.append(str(v))
i, j = i - 1, j - 1
return ''.join(ans[::-1])
class Solution {
public String addBinary(String a, String b) {
StringBuilder sb = new StringBuilder();
for (int i = a.length() - 1, j = b.length() - 1, carry = 0; i >= 0 || j >= 0 || carry > 0;
--i, --j) {
carry += (i >= 0 ? a.charAt(i) - '0' : 0) + (j >= 0 ? b.charAt(j) - '0' : 0);
sb.append(carry % 2);
carry /= 2;
}
return sb.reverse().toString();
}
}
class Solution {
public:
string addBinary(string a, string b) {
string res;
int carry = 0;
int i = a.size() - 1;
int j = b.size() - 1;
while (i >= 0 || j >= 0) {
int digitA = i >= 0 ? a.at(i--) - '0' : 0;
int digitB = j >= 0 ? b.at(j--) - '0' : 0;
int sum = digitA + digitB + carry;
carry = sum >= 2 ? 1 : 0;
sum = sum >= 2 ? sum - 2 : sum;
res += to_string(sum);
}
if (carry == 1) res.push_back('1');
reverse(res.begin(), res.end());
return res;
}
};
using System;
using System.Collections.Generic;
public class Solution {
public string AddBinary(string a, string b) {
var list = new List<char>(Math.Max(a.Length, b.Length) + 1);
var i = a.Length - 1;
var j = b.Length - 1;
var carry = 0;
while (i >= 0 || j >= 0)
{
if (i >= 0)
{
carry += a[i] - '0';
}
if (j >= 0)
{
carry += b[j] - '0';
}
list.Add((char)((carry % 2) + '0'));
carry /= 2;
--i;
--j;
}
if (carry > 0) list.Add((char) (carry + '0'));
list.Reverse();
return new string(list.ToArray());
}
}
func addBinary(a string, b string) string {
for len(a) > len(b) {
b = "0" + b
}
for len(a) < len(b) {
a = "0" + a
}
zero := []byte("0")[0]
ret := make([]byte, len(a))
for right := len(a) - 1; right > 0; right-- {
t := ret[right] + a[right] + b[right] - zero*2
ret[right] = t%2 + zero
if t >= 2 {
ret[right-1] = 1
}
}
t := ret[0] + a[0] + b[0] - zero*2
ret[0] = t%2 + zero
if t >= 2 {
ret = append([]byte("1"), ret...)
}
return string(ret)
}
function addBinary(a: string, b: string): string {
const n = Math.max(a.length, b.length);
const res = [];
let isOver = false;
for (let i = 0; i < n || isOver; i++) {
let val = isOver ? 1 : 0;
isOver = false;
if (a[a.length - i - 1] === '1') {
val++;
}
if (b[b.length - i - 1] === '1') {
val++;
}
if (val > 1) {
isOver = true;
val -= 2;
}
res.push(val);
}
return res.reverse().join('');
}
impl Solution {
pub fn add_binary(a: String, b: String) -> String {
let n = a.len().max(b.len());
let (a, b) = (a.as_bytes(), b.as_bytes());
let mut res = vec![];
let mut is_over = false;
let mut i = 0;
while i < n || is_over {
let mut val = if is_over { 1 } else { 0 };
is_over = false;
if a.get(a.len() - i - 1).unwrap_or(&b'0') == &b'1' {
val += 1;
}
if b.get(b.len() - i - 1).unwrap_or(&b'0') == &b'1' {
val += 1;
}
if val > 1 {
is_over = true;
val -= 2;
}
i += 1;
res.push(char::from(b'0' + val));
}
res.iter().rev().collect()
}
}