408-二进制求和
思路
先相加，在处理进位，为了方便操作，将选字符串翻转，使低位位于数组前端
code
class Solution {
public:
/*
* @param a: a number
* @param b: a number
* @return: the result
*/
string addBinary(string a, string b) {
// write your code here
int sizeA = a.size(), sizeB = b.size();
if (sizeA <= 0 || sizeB <= 0) {
return string();
}
reversalString(a);
reversalString(b);
string result;
int i = 0, carry = 0;
for (i = 0; i<sizeA && i<sizeB; i++) {
result += '0' + (a[i] - '0' + b[i] - '0');
}
for (; i < sizeA; i++) {
result += a[i];
}
for (; i < sizeB; i++) {
result += b[i];
}
for (i = 0; i < result.size(); i++) {
int temp = result[i] - '0' + carry;
if (temp == 2) {
carry = 1;
result[i] = '0';
}
else if (temp == 3) {
carry = 1;
result[i] = '1';
}
else {
carry = 0;
result[i] = temp + '0';
}
}
if (carry == 1) {
result += '1';
}
reversalString(result);
return result;
}
void reversalString(string &str) {
for (int i = 0; i < str.size() / 2; i++) {
swap(str[i], str[str.size() - 1 - i]);
}
}
};