leetcode:67 Add Binary－每日编程第四十三题

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Add Binary

Total Accepted: 67578 Total Submissions: 261262 Difficulty: Easy

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

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

class Solution {public:    void addBinary(string& str,string& a,string& b,int len1,int len2,bool carryBit)    {        if(len1==-1&&len2==-1){            if(carryBit){                str+='1';                return;            }else{                return ;            }        }                if(len1==-1){            if(!carryBit){                char ch=b[len2];                addBinary(str,a,b,len1,len2-1,false);                str+=ch;            }else{                if(b[len2]=='1'){                    char ch='0';                    addBinary(str,a,b,len1,len2-1,true);                    str+=ch;                }else{                    char ch='1';                    addBinary(str,a,b,len1,len2-1,false);                    str+=ch;                }            }        }else if(len2==-1){            if(!carryBit){                char ch=a[len1];                addBinary(str,a,b,len1-1,len2,false);                str+=ch;            }else{                if(a[len1]=='1'){                    char ch='0';                    addBinary(str,a,b,len1-1,len2,true);                    str+=ch;                }else{                    char ch='1';                    addBinary(str,a,b,len1-1,len2,false);                    str+=ch;                }            }        }else{            char ch=a[len1]+b[len2]-'0';            if(ch=='2'){                if(carryBit){                    ch='1';                }else{                    ch='0';                }                addBinary(str,a,b,len1-1,len2-1,true);                str+=ch;            }else if(ch=='1'){                if(carryBit){                    ch='0';                    addBinary(str,a,b,len1-1,len2-1,true);                    str+=ch;                }else{                    addBinary(str,a,b,len1-1,len2-1,false);                    str+=ch;                }            }else{                if(carryBit){                    ch='1';                }                addBinary(str,a,b,len1-1,len2-1,false);                str+=ch;            }        }    }    string addBinary(string a, string b) {        int len1=a.length()-1;        int len2=b.length()-1;        if(a==""){            return b;        }else if(b==""){            return a;        }        string str;        addBinary(str,a,b,len1,len2,false);        return str;    }};

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