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Add 2 Binary Strings #go
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/* | |
Observation: We start from the last char for each string. As long as we have not reached the beginning we compare the chars. The math is | |
1 + 1 = 0 (curry = 1) | |
1 + 0 = 0 + 1 = 1 (curry = 0) | |
1 + 1 = 1 (curry = 1) if curry = 1 | |
1 + 0 = 0 + 1 = 0 (curry = 1) if curry = 1 | |
If we have a curry in the end we prepend the 1 | |
Note: Adding 2 mumbers is similar. We have to calculate: | |
sum := carry + x + y | |
carry = sum / 10 | |
val = sum % 10 | |
*/ | |
func AddBinary(a string, b string) string { | |
res := "" | |
curry := 0 | |
l1 := len(a) - 1 | |
l2 := len(b) - 1 | |
for l1 >= 0 || l2 >= 0 { | |
var left = 0 | |
var right = 0 | |
if l1 >= 0 { | |
if rune(a[l1]) == '1' { | |
left = 1 | |
} | |
l1 -= 1 | |
} | |
if l2 >= 0 { | |
if rune(b[l2]) == '1' { | |
right = 1 | |
} | |
l2 -= 1 | |
} | |
added := curry + left + right | |
switch added { | |
case 0: | |
res = "0" + res | |
curry = 0 | |
case 1: | |
res = "1" + res | |
curry = 0 | |
case 2: | |
res = "0" + res | |
curry = 1 | |
case 3: | |
res = "1" + res | |
curry = 1 | |
} | |
} | |
if curry == 1 { | |
res = "1" + res | |
} | |
return res | |
} |
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