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| 1 | +// |
| 2 | +// Mergesort.swift |
| 3 | +// |
| 4 | +// |
| 5 | +// Created by Kelvin Lau on 2016-02-03. |
| 6 | +// |
| 7 | +// |
| 8 | + |
| 9 | +func mergeSort<T: Comparable>(_ array: [T]) -> [T] { |
| 10 | + guard array.count > 1 else { return array } |
| 11 | + let middleIndex = array.count / 2 |
| 12 | + let leftArray = mergeSort(Array(array[0..<middleIndex])) |
| 13 | + let rightArray = mergeSort(Array(array[middleIndex..<array.count])) |
| 14 | + return merge(leftPile: leftArray, rightPile: rightArray) |
| 15 | +} |
| 16 | + |
| 17 | +func merge<T: Comparable>(leftPile: [T], rightPile: [T]) -> [T] { |
| 18 | + var leftIndex = 0 |
| 19 | + var rightIndex = 0 |
| 20 | + var orderedPile: [T] = [] |
| 21 | + if orderedPile.capacity < leftPile.count + rightPile.count { |
| 22 | + orderedPile.reserveCapacity(leftPile.count + rightPile.count) |
| 23 | + } |
| 24 | + |
| 25 | + while true { |
| 26 | + guard leftIndex < leftPile.endIndex else { |
| 27 | + orderedPile.append(contentsOf: rightPile[rightIndex..<rightPile.endIndex]) |
| 28 | + break |
| 29 | + } |
| 30 | + guard rightIndex < rightPile.endIndex else { |
| 31 | + orderedPile.append(contentsOf: leftPile[leftIndex..<leftPile.endIndex]) |
| 32 | + break |
| 33 | + } |
| 34 | + |
| 35 | + if leftPile[leftIndex] < rightPile[rightIndex] { |
| 36 | + orderedPile.append(leftPile[leftIndex]) |
| 37 | + leftIndex += 1 |
| 38 | + } else { |
| 39 | + orderedPile.append(rightPile[rightIndex]) |
| 40 | + rightIndex += 1 |
| 41 | + } |
| 42 | + } |
| 43 | + |
| 44 | + |
| 45 | + return orderedPile |
| 46 | +} |
| 47 | + |
| 48 | +/* |
| 49 | + This is an iterative bottom-up implementation. Instead of recursively splitting |
| 50 | + up the array into smaller sublists, it immediately starts merging the individual |
| 51 | + array elements. |
| 52 | + |
| 53 | + As the algorithm works its way up, it no longer merges individual elements but |
| 54 | + larger and larger subarrays, until eventually the entire array is merged and |
| 55 | + sorted. |
| 56 | + |
| 57 | + To avoid allocating many temporary array objects, it uses double-buffering with |
| 58 | + just two arrays. |
| 59 | +*/ |
| 60 | +func mergeSortBottomUp<T>(_ a: [T], _ isOrderedBefore: (T, T) -> Bool) -> [T] { |
| 61 | + let n = a.count |
| 62 | + var z = [a, a] // the two working arrays |
| 63 | + var d = 0 // z[d] is used for reading, z[1 - d] for writing |
| 64 | + |
| 65 | + var width = 1 |
| 66 | + while width < n { |
| 67 | + |
| 68 | + var i = 0 |
| 69 | + while i < n { |
| 70 | + |
| 71 | + var j = i |
| 72 | + var l = i |
| 73 | + var r = i + width |
| 74 | + |
| 75 | + let lmax = min(l + width, n) |
| 76 | + let rmax = min(r + width, n) |
| 77 | + |
| 78 | + while l < lmax && r < rmax { |
| 79 | + if isOrderedBefore(z[d][l], z[d][r]) { |
| 80 | + z[1 - d][j] = z[d][l] |
| 81 | + l += 1 |
| 82 | + } else { |
| 83 | + z[1 - d][j] = z[d][r] |
| 84 | + r += 1 |
| 85 | + } |
| 86 | + j += 1 |
| 87 | + } |
| 88 | + while l < lmax { |
| 89 | + z[1 - d][j] = z[d][l] |
| 90 | + j += 1 |
| 91 | + l += 1 |
| 92 | + } |
| 93 | + while r < rmax { |
| 94 | + z[1 - d][j] = z[d][r] |
| 95 | + j += 1 |
| 96 | + r += 1 |
| 97 | + } |
| 98 | + |
| 99 | + i += width*2 |
| 100 | + } |
| 101 | + |
| 102 | + width *= 2 // in each step, the subarray to merge becomes larger |
| 103 | + d = 1 - d // swap active array |
| 104 | + } |
| 105 | + return z[d] |
| 106 | +} |
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