# [67. Add Binary](https://leetcode.com/problems/add-binary) [中文文档](/solution/0000-0099/0067.Add%20Binary/README.md) ## Description <p>Given two binary strings <code>a</code> and <code>b</code>, return <em>their sum as a binary string</em>.</p> <p> </p> <p><strong class="example">Example 1:</strong></p> <pre><strong>Input:</strong> a = "11", b = "1" <strong>Output:</strong> "100" </pre><p><strong class="example">Example 2:</strong></p> <pre><strong>Input:</strong> a = "1010", b = "1011" <strong>Output:</strong> "10101" </pre> <p> </p> <p><strong>Constraints:</strong></p> <ul> <li><code>1 <= a.length, b.length <= 10<sup>4</sup></code></li> <li><code>a</code> and <code>b</code> consist only of <code>'0'</code> or <code>'1'</code> characters.</li> <li>Each string does not contain leading zeros except for the zero itself.</li> </ul> ## Solutions <!-- tabs:start --> ### **Python3** ```python class Solution: def addBinary(self, a: str, b: str) -> str: return bin(int(a, 2) + int(b, 2))[2:] ``` ```python 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]) ``` ### **Java** ```java class Solution { public String addBinary(String a, String b) { var sb = new StringBuilder(); int i = a.length() - 1, j = b.length() - 1; for (int 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(); } } ``` ### **C++** ```cpp class Solution { public: string addBinary(string a, string b) { string ans; int i = a.size() - 1, j = b.size() - 1; for (int carry = 0; i >= 0 || j >= 0 || carry; --i, --j) { carry += (i >= 0 ? a[i] - '0' : 0) + (j >= 0 ? b[j] - '0' : 0); ans.push_back((carry % 2) + '0'); carry /= 2; } reverse(ans.begin(), ans.end()); return ans; } }; ``` ### **Go** ```go func addBinary(a string, b string) string { i, j := len(a)-1, len(b)-1 ans := []byte{} for carry := 0; i >= 0 || j >= 0 || carry > 0; i, j = i-1, j-1 { if i >= 0 { carry += int(a[i] - '0') } if j >= 0 { carry += int(b[j] - '0') } ans = append(ans, byte(carry%2+'0')) carry /= 2 } for i, j := 0, len(ans)-1; i < j; i, j = i+1, j-1 { ans[i], ans[j] = ans[j], ans[i] } return string(ans) } ``` ### **TypeScript** ```ts function addBinary(a: string, b: string): string { return (BigInt('0b' + a) + BigInt('0b' + b)).toString(2); } ``` ```ts function addBinary(a: string, b: string): string { let i = a.length - 1; let j = b.length - 1; let ans: number[] = []; for (let carry = 0; i >= 0 || j >= 0 || carry; --i, --j) { carry += (i >= 0 ? a[i] : '0').charCodeAt(0) - '0'.charCodeAt(0); carry += (j >= 0 ? b[j] : '0').charCodeAt(0) - '0'.charCodeAt(0); ans.push(carry % 2); carry >>= 1; } return ans.reverse().join(''); } ``` ### **Rust** ```rust impl Solution { pub fn add_binary(a: String, b: String) -> String { let mut i = a.len() as i32 - 1; let mut j = b.len() as i32 - 1; let mut carry = 0; let mut ans = String::new(); let a = a.as_bytes(); let b = b.as_bytes(); while i >= 0 || j >= 0 || carry > 0 { if i >= 0 { carry += a[i as usize] - b'0'; i -= 1; } if j >= 0 { carry += b[j as usize] - b'0'; j -= 1; } ans.push_str(&(carry % 2).to_string()); carry /= 2; } ans.chars().rev().collect() } } ``` ### **C#** ```cs public class Solution { public string AddBinary(string a, string b) { int i = a.Length - 1; int j = b.Length - 1; var sb = new StringBuilder(); for (int carry = 0; i >= 0 || j >= 0 || carry > 0; --i, --j) { carry += i >= 0 ? a[i] - '0' : 0; carry += j >= 0 ? b[j] - '0' : 0; sb.Append(carry % 2); carry /= 2; } var ans = sb.ToString().ToCharArray(); Array.Reverse(ans); return new string(ans); } } ``` ### **...** ``` ``` <!-- tabs:end -->