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剑指 Offer II 002. 二进制加法

题目描述

给定两个 01 字符串 a 和 b ，请计算它们的和，并以二进制字符串的形式输出。

输入为 非空 字符串且只包含数字 1 和 0。

 

示例 1:

输入: a = "11", b = "10"
输出: "101"

示例 2:

输入: a = "1010", b = "1011"
输出: "10101"

 

提示：

  • 每个字符串仅由字符 '0' 或 '1' 组成。
  • 1 <= a.length, b.length <= 10^4
  • 字符串如果不是 "0" ，就都不含前导零。

 

注意：本题与主站 67 题相同：https://leetcode.cn/problems/add-binary/

解法

方法一：模拟

我们用一个变量 $carry$ 记录当前的进位，用两个指针 $i$ 和 $j$ 分别指向 $a$ 和 $b$ 的末尾，从末尾到开头逐位相加即可。

时间复杂度 $O(\max(m, n))$，其中 $m$ 和 $n$ 分别为字符串 $a$ 和 $b$ 的长度。空间复杂度 $O(1)$。

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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])

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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();
    }
}

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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;
    }
};

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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)
}

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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('');
}

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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()
    }
}

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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);
    }
}

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class Solution {
    func addBinary(_ a: String, _ b: String) -> String {
        var result = ""
        var carry = 0
        var i = a.count - 1, j = b.count - 1

        let aChars = Array(a)
        let bChars = Array(b)

        while i >= 0 || j >= 0 || carry > 0 {
            let digitA = i >= 0 ? Int(String(aChars[i]))! : 0
            let digitB = j >= 0 ? Int(String(bChars[j]))! : 0

            carry += digitA + digitB
            result = "\(carry % 2)" + result
            carry /= 2

            i -= 1
            j -= 1
        }

        return result
    }
}

方法二

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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])

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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('');
}

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