README_EN.md

comments	difficulty	edit_url	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/0000-0099/0067.Add%20Binary/README_EN.md	Bit Manipulation Math String Simulation

67. Add Binary

中文文档

Description

Given two binary strings a and b, return their sum as a binary string.

 

Example 1:

Input: a = "11", b = "1"
Output: "100"

Example 2:

Input: a = "1010", b = "1011"
Output: "10101"

 

Constraints:

• 1 <= a.length, b.length <= 104
• a and b consist only of '0' or '1' characters.
• Each string does not contain leading zeros except for the zero itself.

Solutions

Solution 1: Simulation

We use a variable $carry$ to record the current carry, and two pointers $i$ and $j$ to point to the end of $a$ and $b$ respectively, and add them bit by bit from the end to the beginning.

The time complexity is $O(\max(m, n))$, where $m$ and $n$ are the lengths of strings $a$ and $b$ respectively. The space complexity is $O(1)$.

Python3

class Solution:
    def addBinary(self, a: str, b: str) -> str:
        return bin(int(a, 2) + int(b, 2))[2:]

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++

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

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

function addBinary(a: string, b: string): string {
    return (BigInt('0b' + a) + BigInt('0b' + b)).toString(2);
}

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#

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);
    }
}

Solution 2

Python3

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])

TypeScript

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('');
}