leetcode 67 Add Binary

Add Binary Total Accepted: 46815 Total Submissions: 189215 My Submissions

Given two binary strings, return their sum (also a binary string).

For example, a = "11" b = "1" Return "100".

我的解决方案：

代码语言：javascript复制

class Solution {
public:
    string addBinary(string a, string b) 
    {
        
        string result = "";
        int c = 0;
        int i = a.size() - 1;
        int j = b.size() - 1;
        
        while(i >= 0 || j >=0 ||c ==1)
        {
            c  = i >= 0 ? a[i--] - '0':0;
            c  = j >= 0 ? b[j--] - '0':0;
            result = char( c% 2   '0')   result;
            c /= 2;
            
        }
        
        return result;
        
        
    }
};

c语言解决方案：

代码语言：javascript复制

char* addBinary(char* a, char* b) {
    int n, m;
    for (n=0; *a; a  , n  ) ;
    for (m=0; *b; b  , m  ) ;
    char *p = (char*)malloc(m>n ? m 2 : n 2), *last = p;
    int c = 0;
    while (n || m || c) {
        int s = c;
        if (n) {
            s  = *(--a)-'0';
            --n;
        }
        if (m) {
            s  = *(--b)-'0';
            --m;
        }
        *last   = (s&1) '0';
        c = s>>1;
    }
    *last=0;
    char *start = p, t;
    while (start 1 < last) { // reverse string
        t = *start;
        *start  =*(--last);
        *last=t;
    }
    return p;
}

数字电路版本代码：加法器的实现 https://leetcode.com/discuss/40846/c-4ms-solution-inspired-by-hardware-full-adder-circuit

python 的三个版本：

代码语言：javascript复制

class Solution:
    # @param {string} a
    # @param {string} b
    # @return {string}
    def addBinary(self, a, b):
        i, m, n, result, carry = 1, len(a), len(b), [], 0
        while i <= m or i <= n:
            temp = carry
            if i <= m:
                temp  = int(a[-i])
            if i <= n:
                temp  = int(b[-i])

            carry = temp / 2
            result.append(str(temp % 2))
            i  = 1

        if carry:
            result.append(str(carry))

        return ''.join(result[::-1])

or a really short one if you want

class Solution:
    # @param {string} a
    # @param {string} b
    # @return {string}
    def addBinary(self, a, b):
        return '{0:b}'.format(int(a, 2)   int(b, 2))


class Solution:
    # @param {string} a
    # @param {string} b
    # @return {string}
    def addBinary(self, a, b):
        return bin(int(a,2)   int(b,2))[2:]

python add binary return string

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