67. binary sum

Topic introduction ：

Here are two binary strings , Returns the sum of them （ In binary ）.

Input is Non empty String and contains only numbers 1 and 0.

Example ：

 Input : a = "11", b = "1"

 Output : "100"

Then we have two solutions to this problem ：

1. The string a ,b The binary represented in is converted into numbers , Add the two to get the sum , Finally, the binary of the sum is stored in a return array .

2. We do a similar addition operation for two strings , Last result returned .

about 1 Although the method is the easiest to understand , But it also faces the problem of overflow in different situations , So here we use the second method to realize the problem .

The code is as follows ：

char * addBinary(char * a, char * b){

    int Base = 0;	// carry 

	int length = (strlen(a)>strlen(b)? strlen(a)+2:strlen(b)+2);  
    //+2 The reason for this is that the two binaries add up to one bit , The other is used to fill in '\0'

	char* result = (char*)malloc(sizeof(char)*length);	// Open up space 

	result[length-1] = '\0';   // The last person to open up space is '\0' It means the ending 

	for(int i = strlen(a)-1,j = strlen(b)-1,k = length -2; (i >=0)||(j >= 0); i--,j--,k--)
	{
		int sum = Base;
		sum += (i >= 0? a[i]-'0':0);
		sum += (j >= 0? b[j]-'0':0);
		Base = sum /2;
		result[k] = '0'+ sum % 2;
	}
	if(Base == 0)  // No carry , Go straight back to 
		return result+1;    // Last +1  Then skip the address without carry , But it may lead to cross-border 
	result[0] = '1';    // Have carry , Make up for the highest position 
		return result;
}