Leetcode 67. 二进制求和 C++

题目描述

方法一

暴力法

class Solution {
public:
    string addBinary(string a, string b) {
        string res;
        if(a.empty() && b.empty()) return res;
        if(a.empty() || b.empty()) return a.empty()? b:a;
        int carry=0;
        if(a.size() < b.size()) swap(a,b);//保证a是较长的那一个
        auto a_begin=a.begin()-1;
        auto a_end  = a.end()-1;
        auto b_begin=b.begin()-1;
        auto b_end=b.end()-1;
        while(a_end!=a_begin && b_end!=b_begin)
        {
            if( (*a_end-'0')+(*b_end-'0')+carry == 0) 
            {
                res='0'+res;
                carry=0;
            }
            else if( (*a_end-'0')+(*b_end-'0')+carry == 1)
            {
                res='1'+res;
                carry=0;
            }
            else if( (*a_end-'0')+(*b_end-'0')+carry == 2 )
            {
                res='0'+res;
                carry=1;
            }
            else if( (*a_end-'0')+(*b_end-'0')+carry == 3)
            {
                res='1'+res;
                carry=1;
            }
            --a_end;
            --b_end;
        }
        while(a_begin!=a_end)
        {
            if(carry+(*a_end-'0')==0)
            {
                 res='0'+res;
                carry=0;
            }
            else if(carry+(*a_end-'0')==1)
            {
                res='1'+res;
                carry=0;
            }
            else if(carry+(*a_end-'0')==2)
            {
                res='0'+res;
                carry=1;
            }
            else if(carry+ (*a_end-'0')==3)
            {
                res='1'+res;
                carry=1;
            }
            --a_end;
        }
        if(carry==1)
            res='1'+res;
        return res;
    }
};

方法二

该方法参考自 ：https://www.cnblogs.com/ariel-dreamland/p/9151374.html
这种方法比较简洁，不需要考虑两个字符串的长度。用到的思想是补零的思想，将较短的字符串前面补零，之后长度相同就可以了。

class Solution {
public:
    string addBinary(string a, string b) {
        string res;
        int carry=0;
        int m=a.size()-1, n=b.size()-1;
        while( m>=0 || n>=0 )
        {
            int q = (m>=0)? a[m--]-'0':0;
            int p = (n>=0)? b[n--]-'0':0;
            int k=q+p+carry;
            res= to_string(k%2)+res; //这里需要用到to_string,(k%2)+'0是错误的，这种加法操作只适用于字符串常量，这里k是变量
            carry=k/2;
        }
        if(1 == carry) res='1'+res;
        return res;
    }
};