implementation by Swift
Given an array of n integers where n > 1, nums, return an array output such that output[i] is equal to the product of all the elements of nums except nums[I]. Solve it without division and in O(n). For example, given [1,2,3,4], return [24,12,8,6].
[ 1, a[0], a[0]*a[1], a[0]*a[1]*a[2], ] [ a[1]*a[2]*a[3], a[2]*a[3], a[3], 1, ]
func productArray(inputArray: Array<Int>) -> Array<Int> {
guard !inputArray.isEmpty else {
// throws
return []
}
let arrayCount = inputArray.count
var outPutArray = Array(repeating: 1, count: arrayCount)
for n in 1...(arrayCount - 1) {
outPutArray[n] = outPutArray[n - 1] * inputArray[n - 1]
}
var right = 1
for n in 0...(arrayCount - 1) {
let index = arrayCount - 1 - n
outPutArray[index] *= right
right *= inputArray[index]
}
debugPrint("inputArray ---- \(inputArray); outPutArray --- \(outPutArray);")
return outPutArray
}
Given an array nums, write a function to move all non-zero numbers to the head of it while maintaining the relative order of the non-zero numbers. For example, given nums = [0, 1, 0, 3, 12], after calling your function, nums should be [1, 3, 12, 0, 0]. Note: You must do this in-place without making a copy of the array. Minimize the total number of operations.
// MARK: - 2. Number first
func NumberFirst(input: Array<Int>) -> Array<Int> {
var inputArray = input
guard !inputArray.isEmpty else {
// throws
return []
}
let arrayCount = inputArray.count
var num = 0
var zeroCount = 0
for n in 0...(arrayCount - 1) {
if inputArray[n] == 0 {
zeroCount += 1;
} else {
inputArray[num] = inputArray[n]
num += 1
}
}
for _ in 0...(zeroCount - 1) {
inputArray[num] = 0
num += 1
}
debugPrint("NumberFirst ---- \(inputArray)")
return inputArray
}
Implement pow(x, n).
func Power(num1:Double, num2:Int) -> Double {
let x = num1
var n = num2
if x == 0 && n == 0 {
// throws
return 0
}
var isNegative = false
if n < 0 {
n *= -1
isNegative = true
}
let result = pow(x: x, n: n)
if isNegative {
return 1.0 / result
} else {
return result
}
}
func pow(x:Double, n:Int) -> Double {
if n == 0 {
return 1
} else {
let result = pow(x: x, n: n / 2)
if n % 2 != 0 {
return x * result * result
} else {
return result * result
}
}
}
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