Add Binary

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

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

 

类似模拟十进制加法

 1 class Solution {
 2 public:
 3     string addBinary(string a, string b) {
 4         if( a.empty() ) return b;
 5         if( b.empty() ) return a;
 6         string ans;
 7         reverse(a.begin(), a.end());    //反转后，从低位开始相加
 8         reverse(b.begin(), b.end());
 9         int i=0;
10         int carry = 0;
11         while( i<a.length() && i<b.length() ) { //处理他们共同都有的部分
12             int sum = a[i] - '0' + b[i] - '0' + carry;
13             if( sum == 3 ) {
14                 ans.push_back('1');
15                 carry = 1;
16             } else if( sum == 2 ) {
17                 ans.push_back('0');
18                 carry = 1;
19             } else if( sum == 1 ) {
20                 ans.push_back('1');
21                 carry = 0;
22             } else if( sum == 0 ) {
23                 ans.push_back('0');
24                 carry = 0;
25             }
26             ++i;
27         }
28         string& c = a.length() > b.length() ? a : b;
29         while( i<c.length() ) { //处理一方多余的部分，及进位
30             int sum = c[i] - '0' + carry;
31             if( sum == 2 ) {
32                 ans.push_back('0');
33                 carry = 1;
34             } else if( sum == 1 ) {
35                 ans.push_back('1');
36                 carry = 0;
37             } else if( sum == 0 ) {
38                 ans.push_back('0');
39                 carry = 0;
40             }
41             ++i;
42         }
43         if( carry ) ans.push_back('1'); //最后若有进位
44         reverse(ans.begin(), ans.end());
45         return ans;
46     }
47 };

 

class Solution {
public:
    string addBinary(string a, string b) {
        return addStringBaseN(a, b, 2);
    }
    
    string addStringBaseN(string num1, string num2, int base) {
        std::reverse(num1.begin(), num1.end());
        std::reverse(num2.begin(), num2.end());
        string sum_string;
        int i = 0;
        int carry = 0;
        while (i < num1.size() && i < num2.size()) {
            int sum = (num1[i] - '0') + (num2[i] - '0') + carry;
            sum_string.push_back((sum % base) + '0');
            carry = sum / base;
            ++i;
        }
        
        const string& lstring = num1.size() > num2.size() ? num1 : num2;    
        while (i < lstring.size()) {
            int sum = lstring[i] - '0' + carry;
            sum_string.push_back(sum % base + '0');
            carry = sum / base;
            ++i;
        }
        
        if (carry > 0) {
            sum_string.push_back(carry + '0');
        }
        std::reverse(sum_string.begin(), sum_string.end());
        return sum_string;
    }
};

 

posted on 2014-10-06 21:57  bug睡的略爽  阅读(145)  评论(0)  编辑  收藏  举报

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