題目：

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

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

分析：可以根據(jù)二進(jìn)制的加法原理解決此問題，要注意string與int之間的轉(zhuǎn)換。

代碼如下：

??????? string addBinary(string a, string b) {
??????? int n1=a.length();
??????? if(n1==0)return b;
??????? int n2=b.length();
??????? if(n2==0)return a;
??????? string result;
??????? int t=0;
??????? if(n1<n2)
??????? {
??????????? string tmp(a);
??????????? a=b;
??????????? b=tmp;
??????????? int temp=n1;
??????????? n1=n2;
??????????? n2=temp;
??????? }
??????? for(int i=n2-1;i>=0;i--)
??????? {
?????????? ?
??????????? string tmp;
??????????? tmp='0'+(int(a[n1-1]-'0')+int(b[i]-'0')+t)%2; //此句中 int(a[n1-1]-'0') 是string轉(zhuǎn)int

??????????????????????????????????????????????????????????????????????????????????????? //int + '0' 是int 轉(zhuǎn)化為string
??????????? result=tmp+result;
??????????? t=(int(a[n1-1]-'0')+int(b[i]-'0')+t)/2;
??????????? n1--;
??????? }
??????? for(int i=n1-1;i>=0;i--)
??????? {
??????????? string tmp;
??????????? tmp='0'+(int(a[i]-'0')+t)%2;
??????????? result=tmp+result;
??????????? t=(int(a[i]-'0')+t)/2;
??????? }
??????? if(t==1)
??????? {
??????????? string tmp;
??????????? tmp='0'+1;
??????????? result=tmp+result;
??????????? return result;
??????? }
??????? return result;
??? }

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