Give an algorithm that determines the number of inversions in any permutation on n elements in O(nlgn) worst-case time.
///Merge Sort. Θ(nlgn)
func mergeInversionsSort<T: Comparable>(array: inout [T]) -> Int {
return _mergeInversionsSort(array: &array, lower: 0, upper: array.count-1);
}
///Inner representation
func _mergeInversionsSort<T: Comparable>(array: inout [T], lower: Int, upper: Int) -> Int {
guard lower < upper else { return 0 }
//Note:upper+lower NOT upper-lower
let divider = (upper+lower)/2
var count = 0
count += _mergeInversionsSort(array: &array, lower: lower, upper: divider)
count += _mergeInversionsSort(array: &array, lower: divider+1, upper: upper)
count += mergeInversions(of: &array, lower: lower, divider: divider, upper: upper)
return count;
}
///Merge Inversion. Θ(n)
func mergeInversions<T: Comparable>(of array:inout [T], lower:Int, divider:Int, upper:Int) -> Int {
var count = 0
let prefix = Array(array[lower...divider])
let suffix = Array(array[divider+1...upper])
//Note:k = lower NOT k = 0
var i = 0, j = 0, k = lower
while i<prefix.count, j<suffix.count {
if prefix[i] <= suffix[j] {
array[k] = prefix[i]
i += 1
} else {
array[k] = suffix[j]
j += 1
//Core inversions statistics
count += prefix.count-i
}
k += 1
}
if i == prefix.count {
while j < suffix.count {
array[k] = suffix[j]
k += 1
j += 1
}
} else {
while i < prefix.count {
array[k] = prefix[i]
k += 1
i += 1
}
}
return count
}
var s = try Int.randomArray()
mergeInversionsSort(array: &s)
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