67.Add Binary(String-Easy)

转载请注明作者和出处： http://blog.csdn.net/c406495762

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

For example,

a = “11”

b = “1”

Return “100”

题目：两个代表二进制数字的字符串相加

思路：从低位开始相加，逢二进一：

- 字符串的高位是数字的低位。例如：string a = “100”，a[2]则为”0”； - 用一个变量记录相加的进位情况。 > Language：cpp

class Solution {
public:
    string addBinary(string a, string b) {
        string s = "";
        //i字符串a的索引值,j字符串b的索引值
        int c = 0, i = a.size() - 1, j = b.size() - 1;
        //字符串a和b没有遍历完，或者存在进位，继续计算
        while(i >= 0 || j >= 0 || c == 1)
        {
            //三目运算符：低位相加
            c += i >= 0 ? a[i--] - '0' : 0;
            c += j >= 0 ? b[j--] - '0' : 0;
            //结果
            s = char(c % 2 + '0') + s;
            //进位计算
            c /= 2;
        }
        //返回结果
        return s;
    }
};

Language:Python

class Solution(object):
    def addBinary(self, a, b):
        """
        :type a: str
        :type b: str
        :rtype: str
        """
        if len(a)==0: return b
        if len(b)==0: return a
        if a[-1] == '1' and b[-1] == '1':
            return self.addBinary(self.addBinary(a[0:-1],b[0:-1]),'1') +'0'
        if a[-1] == '0' and b[-1] == '0':
            return self.addBinary(a[0:-1],b[0:-1])+'0'
        else:
            return self.addBinary(a[0:-1],b[0:-1])+'1'

LeetCode题目汇总： https://github.com/Jack-Cherish/LeetCode