英文:
How to print each digit of the number different from the one that precedes it?
问题
例如，如果我有数字35565，输出结果是3565。
所以，我的代码片段获取了每个数字，但我不知道如何保留前一个数字以便与下一个数字进行比较。
for {
num = num / 10
fmt.Print(num)
if num/10 == 0 {
break
}
}
英文:
For example, if I have the number 35565, the output is 3565.
So, my snippet gets single digits, but I don't know how to keep previous digit to check it with the next one.
for {
num = num / 10
fmt.Print(num)
if num/10 == 0 {
break
}
}
答案1
得分: 0
这种方法从右到左将数字分解为其各个位数，并将它们存储为整数切片，然后从左到右迭代这些位数，以构建具有“顺序唯一”位数的数字。
我最初尝试从左到右分解数字，但无法解决如何处理占位零；从右到左分解，我知道如何捕获这些零。
// unique 从非负数 x 中移除重复数字序列，
// 仅返回“顺序唯一”的数字：
// 12→12, 122→12, 1001→101, 35565→3565.
//
// 负数 x 返回 -1。
func unique(x int) int {
switch {
case x < 0:
return -1
case x <= 10:
return x
}
// -- 将 x 分解为其各个位数
var (
mag int // x 的数量级
nDigits int // x 的位数
digits []int // x 的各个位数
)
mag = int(math.Floor(math.Log10(float64(x))))
nDigits = mag + 1
// 从右到左工作以保留占位零
digits = make([]int, nDigits)
for i := nDigits - 1; i >= 0; i-- {
digits[i] = x % 10
x /= 10
}
// -- 从左到右构建新的“顺序唯一”x
var prevDigit, newX int
for _, digit := range digits {
if digit != prevDigit {
newX = newX*10 + digit
}
prevDigit = digit
}
return newX
}
这里有一个带有测试的Go Playground。
可以通过在开头翻转负号并在结尾恢复来适应处理负数。
英文:
This approach breaks down a number into its digits from right-to-left, storing them as a slice of ints, then iterates those digits from left-to-right to build up the number with "sequentially unique" digits.
I originally tried to break down the number from left-to-right but couldn't figure out how to handle place-holding zeroes; breaking it down from right-to-left, I know how to capture those zeroes.
// 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
}
Here's a Go Playground with a test.
This could be adapted to deal with negative numbers by flipping the negative sign at the beginning and restoring it at the end.
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