javascript 解法, 执行用时: 88 ms, 内存消耗: 43.1 MB, 提交时间: 2023-10-28 10:58:46

/**
 * @param {string} a
 * @param {string} b
 * @return {string}
 */
var addBinary = function(a, b) {
    let ans = "";
    let ca = 0;
    for(let i = a.length - 1, j = b.length - 1;i >= 0 || j >= 0; i--, j--) {
        let sum = ca;
        sum += i >= 0 ? parseInt(a[i]) : 0;
        sum += j >= 0 ? parseInt(b[j]) : 0;
        ans += sum % 2;
        ca = Math.floor(sum / 2);
    }
    ans += ca == 1 ? ca : "";
    return ans.split('').reverse().join('');
};

golang 解法, 执行用时: 0 ms, 内存消耗: 3.4 MB, 提交时间: 2023-10-28 10:58:21

func addBinary(a string, b string) string {
    ans := ""
    carry := 0
    lenA, lenB := len(a), len(b)
    n := max(lenA, lenB)

    for i := 0; i < n; i++ {
        if i < lenA {
            carry += int(a[lenA-i-1] - '0')
        }
        if i < lenB {
            carry += int(b[lenB-i-1] - '0')
        }
        ans = strconv.Itoa(carry%2) + ans
        carry /= 2
    }
    if carry > 0 {
        ans = "1" + ans
    }
    return ans
}

func max(x, y int) int {
    if x > y {
        return x
    }
    return y
}

cpp 解法, 执行用时: 0 ms, 内存消耗: 6.6 MB, 提交时间: 2023-10-28 10:58:06

class Solution {
public:
    string addBinary(string a, string b) {
        string ans;
        reverse(a.begin(), a.end());
        reverse(b.begin(), b.end());

        int n = max(a.size(), b.size()), carry = 0;
        for (size_t i = 0; i < n; ++i) {
            carry += i < a.size() ? (a.at(i) == '1') : 0;
            carry += i < b.size() ? (b.at(i) == '1') : 0;
            ans.push_back((carry % 2) ? '1' : '0');
            carry /= 2;
        }

        if (carry) {
            ans.push_back('1');
        }
        reverse(ans.begin(), ans.end());

        return ans;
    }
};

python3 解法, 执行用时: 44 ms, 内存消耗: 15.9 MB, 提交时间: 2023-10-28 10:57:52

class Solution:
    def addBinary(self, a, b) -> str:
        x, y = int(a, 2), int(b, 2)
        while y:
            answer = x ^ y
            carry = (x & y) << 1
            x, y = answer, carry
        return bin(x)[2:]

java 解法, 执行用时: 1 ms, 内存消耗: 40.1 MB, 提交时间: 2023-10-28 10:57:33

class Solution {
    public String addBinary1(String a, String b) {
        return Integer.toBinaryString(
            Integer.parseInt(a, 2) + Integer.parseInt(b, 2)
        );
    }

    public String addBinary(String a, String b) {
        StringBuffer ans = new StringBuffer();

        int n = Math.max(a.length(), b.length()), carry = 0;
        for (int i = 0; i < n; ++i) {
            carry += i < a.length() ? (a.charAt(a.length() - 1 - i) - '0') : 0;
            carry += i < b.length() ? (b.charAt(b.length() - 1 - i) - '0') : 0;
            ans.append((char) (carry % 2 + '0'));
            carry /= 2;
        }

        if (carry > 0) {
            ans.append('1');
        }
        ans.reverse();

        return ans.toString();
    }
}

python3 解法, 执行用时: 56 ms, 内存消耗: N/A, 提交时间: 2018-09-21 16:24:28

class Solution:
    def addBinary(self, a, b):
        """
        :type a: str
        :type b: str
        :rtype: str
        """
        return bin(int(a, 2) + int(b, 2))[2:]