AddBinary

Add Binary

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

For example,

a = "11"
b = "1"
Return "100". 

Solution

从字符串末尾逐一相加，直到其中一个字符串结束时，若还有字符串，则追加到结果

注意处理进位，尤其不要丢掉最后一位进位

class Solution {
	public:
		string addBinary(string a, string b) {
			int i = a.size() - 1;
			int j = b.size() - 1; 
			string result;
			char carray = '0';
			while (i >= 0 && j >= 0) {
				char c = add(a[i], b[j], carray);
				switch (c) {
					case '0':
					case '1':
						carray = '0';
						result.push_back(c);
						break;
					case '2':
						carray = '1';
						result.push_back('0');
						break;
					case '3':
						carray = '1';
						result.push_back('1');
						break;
				}
				i--, j--;
			}
			while (i >= 0) {
				char c = add(a[i], carray);
				if (c == '1' || c == '0') {
					result.push_back(c);
					carray = '0';
				} else {
					result.push_back('0');
					carray = '1';
				}
				i--;
			}
			while (j >= 0) {
				char c = add(b[j], carray);
				if (c == '1' || c == '0') {
					result.push_back(c);
					carray = '0';
				} else {
					result.push_back('0');
					carray = '1';
				}
				j--;
			}
			if (carray == '1') { // 最后一位进位
				result.push_back('1');
			}
			return string(result.rbegin(), result.rend());
		}
	private:
		char add(char a, char b) {
			return a + b - '0';
		}
		char add(char a, char b, char c) {
			return a + b + c - '0' - '0';
		}
};