How to print a number that is different from the previous number?

Question content

For example, if I have the number 35565, the output is 3565.

So, my code snippet gets the single digits, but I don't know how to keep the previous number to check with the next number.

for {
num = num / 10
fmt.Print(num)
        
if num/10 == 0 {
    break
}
}

Correct answer

This method breaks the number into numbers from right to left, stores them as integer slices, and then iterates over the numbers from left to right to Construct numbers with "consecutively unique" digits.

I initially tried breaking the number from left to right, but didn't know how to handle the placeholder zeros; breaking it up right to left, I knew how to capture those zeros.

// unique removes sequences of repeated digits from non-negative x,
// returning only "sequentially unique" digits:
// 12→12, 122→12, 1001→101, 35565→3565.
//
// Negative x yields -1.
func unique(x int) int {
    switch {
    case x < 0:
        return -1
    case x <= 10:
        return x
    }

    // -- Split x into its digits
    var (
        mag     int   // the magnitude of x
        nDigits int   // the number of digits in x
        digits  []int // the digits of x
    )

    mag = int(math.Floor(math.Log10(float64(x))))
    nDigits = mag + 1

    // work from right-to-left to preserve place-holding zeroes
    digits = make([]int, nDigits)
    for i := nDigits - 1; i >= 0; i-- {
        digits[i] = x % 10
        x /= 10
    }

    // -- Build new, "sequentially unique", x from left-to-right
    var prevDigit, newX int

    for _, digit := range digits {
        if digit != prevDigit {
            newX = newX*10 + digit
        }
        prevDigit = digit
    }

    return newX
}

This is a go playground for testing.

Can be adapted to handle negative numbers by flipping the negative sign at the beginning and restoring it at the end.

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