代碼倉庫
代碼倉庫包含了完整的代碼實現和測試,其中 Java 版本是官方實現,其他語言版本來自社區貢獻。對於每一種排序算法,如果有多種實現方法,都會盡量提供。另外代碼倉庫中還提供了完備的測試用例,以此確保實現的排序算法覆蓋了每一種可能的情況。
GitHub 鏈接:https://github.com/HawsteinStudio/algocasts-sorting-algorithms
冒泡排序
視頻鏈接:https://algocasts.io/series/sorting-algorithms/episodes/AEpoDvWQ
截圖:
代碼:
public class BubbleSort {
private void swap(int[] arr, int i, int j) {
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
}
// Time: O(n^2), Space: O(1)
public void sort(int[] arr) {
if (arr == null || arr.length == 0) return;
int n = arr.length;
for (int end = n-1; end > 0; --end) {
for (int i = 0; i < end; ++i) {
if (arr[i] > arr[i+1]) {
int tmp = arr[i];
arr[i] = arr[i+1];
arr[i+1] = tmp;
}
}
}
}
// Time: O(n^2), Space: O(1)
public void sortShort(int[] arr) {
if (arr == null || arr.length == 0) return;
int n = arr.length;
for (int end = n-1; end > 0; --end)
for (int i = 0; i < end; ++i)
if (arr[i] > arr[i+1])
swap(arr, i, i+1);
}
// Time: O(n^2), Space: O(1)
public void sortEarlyReturn(int[] arr) {
if (arr == null || arr.length == 0) return;
int n = arr.length;
boolean swapped;
for (int end = n-1; end > 0; --end) {
swapped = false;
for (int i = 0; i < end; ++i) {
if (arr[i] > arr[i+1]) {
swap(arr, i, i+1);
swapped = true;
}
}
if (!swapped) return;
}
}
// Time: O(n^2), Space: O(1)
public void sortSkip(int[] arr) {
if (arr == null || arr.length == 0) return;
int n = arr.length;
int newEnd;
for (int end = n-1; end > 0;) {
newEnd = 0;
for (int i = 0; i < end; ++i) {
if (arr[i] > arr[i+1]) {
swap(arr, i, i+1);
newEnd = i;
}
}
end = newEnd;
}
}
}
選擇排序
視頻鏈接:https://algocasts.io/series/sorting-algorithms/episodes/Z5mzdwpd
截圖:
代碼:
public class SelectionSort {
private void swap(int[] arr, int i, int j) {
if (i == j) return;
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
}
// Time: O(n^2), Space: O(1)
public void sort(int[] arr) {
if (arr == null || arr.length == 0) return;
int n = arr.length;
for (int i = 0; i < n; ++i) {
int minIdx = i;
for (int j = i+1; j < n; ++j)
if (arr[j] < arr[minIdx])
minIdx = j;
swap(arr, i, minIdx);
}
}
// Time: O(n^2), Space: O(1)
public void sortFromEnd(int[] arr) {
if (arr == null || arr.length == 0) return;
int n = arr.length;
for (int i = n-1; i > 0; --i) {
int maxIdx = i;
for (int j = 0; j < i; ++j)
if (arr[j] > arr[maxIdx])
maxIdx = j;
swap(arr, i, maxIdx);
}
}
}
插入排序
視頻鏈接:https://algocasts.io/series/sorting-algorithms/episodes/dbGY9eG5
截圖:
代碼:
public class InsertionSort {
// Time: O(n^2), Space: O(1)
public void sort(int[] arr) {
if (arr == null || arr.length == 0) return;
for (int i = 1; i < arr.length; ++i) {
int cur = arr[i];
int j = i - 1;
while (j >= 0 && arr[j] > cur) {
arr[j+1] = arr[j];
--j;
}
arr[j+1] = cur;
}
}
}
希爾排序
視頻鏈接:https://algocasts.io/series/sorting-algorithms/episodes/zbmKZgWZ
截圖:
代碼:
public class ShellSort {
// Time: O(n^2), Space: O(1)
public void sort(int[] arr) {
if (arr == null || arr.length == 0) return;
for (int gap = arr.length>>1; gap > 0; gap >>= 1) {
for (int i = gap; i < arr.length; ++i) {
int cur = arr[i];
int j = i - gap;
while (j >= 0 && arr[j] > cur) {
arr[j+gap] = arr[j];
j -= gap;
}
arr[j+gap] = cur;
}
}
}
// Time: O(n^2), Space: O(1)
public void insertionSort(int[] arr) {
if (arr == null || arr.length == 0) return;
for (int i = 1; i < arr.length; ++i) {
int cur = arr[i];
int j = i - 1;
while (j >= 0 && arr[j] > cur) {
arr[j + 1] = arr[j];
j -= 1;
}
arr[j + 1] = cur;
}
}
}
快速排序
視頻鏈接:https://algocasts.io/series/sorting-algorithms/episodes/kVG9Pxmg
截圖:
代碼:
public class QuickSort {
private void swap(int[] arr, int i, int j) {
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
}
// [ ... elem < pivot ... | ... elem >= pivot ... | unprocessed elements ]
// i j
private int lomutoPartition(int[] arr, int low, int high) {
int pivot = arr[high];
int i = low;
for (int j = low; j < high; ++j) {
if (arr[j] < pivot) {
swap(arr, i, j);
++i;
}
}
swap(arr, i, high);
return i;
}
// lomuto partition 的另一種實現,可以把最後的 swap 合併到循環中。
private int lomutoPartition2(int[] arr, int low, int high) {
int pivot = arr[high];
int i = low;
for (int j = low; j <= high; ++j) {
if (arr[j] <= pivot) {
swap(arr, i, j);
++i;
}
}
return i-1;
}
private void lomutoSort(int[] arr, int low, int high) {
if (low < high) {
int k = lomutoPartition(arr, low, high);
lomutoSort(arr, low, k-1);
lomutoSort(arr, k+1, high);
}
}
private int hoarePartitionDoWhile(int[] arr, int low, int high) {
int pivot = arr[low + (high-low)/2];
int i = low-1, j = high+1;
while (true) {
do {
++i;
} while (arr[i] < pivot);
do {
--j;
} while (arr[j] > pivot);
if (i >= j) return j;
swap(arr, i, j);
}
}
// [ ... elem <= pivot ... | unprocessed elements | ... elem >= pivot ... ]
// i j
private int hoarePartition(int[] arr, int low, int high) {
int pivot = arr[low + (high-low)/2];
int i = low, j = high;
while (true) {
while (arr[i] < pivot) ++i;
while (arr[j] > pivot) --j;
if (i >= j) return j;
swap(arr, i++, j--);
}
}
private void hoareSort(int[] arr, int low, int high) {
if (low < high) {
int k = hoarePartition(arr, low, high);
hoareSort(arr, low, k);
hoareSort(arr, k+1, high);
}
}
// Time: O(n*log(n)), Space: O(n)
public void lomutoSort(int[] arr) {
if (arr == null || arr.length == 0) return;
lomutoSort(arr, 0, arr.length-1);
}
// Time: O(n*log(n)), Space: O(n)
public void hoareSort(int[] arr) {
if (arr == null || arr.length == 0) return;
hoareSort(arr, 0, arr.length-1);
}
}
歸併排序
視頻鏈接:https://algocasts.io/series/sorting-algorithms/episodes/M0G2k7pz
截圖:
代碼:
public class MergeSort {
// sorted sub-array 1: arr[low ... mid]
// sorted sub-array 2: arr[mid+1 ... high]
private void merge(int[] arr, int low, int mid, int high, int[] tmp) {
int i = low, j = mid + 1, k = 0;
while (i <= mid && j <= high) {
if (arr[i] <= arr[j]) tmp[k++] = arr[i++];
else tmp[k++] = arr[j++];
}
while (i <= mid) tmp[k++] = arr[i++];
while (j <= high) tmp[k++] = arr[j++];
System.arraycopy(tmp, 0, arr, low, k);
}
private void mergeSort(int[] arr, int low, int high, int[] tmp) {
if (low < high) {
int mid = low + (high - low) / 2;
mergeSort(arr, low, mid, tmp);
mergeSort(arr, mid + 1, high, tmp);
merge(arr, low, mid, high, tmp);
}
}
// Time: O(n*log(n)), Space: O(n)
public void sortRecursive(int[] arr) {
if (arr == null || arr.length == 0) return;
int[] tmp = new int[arr.length];
mergeSort(arr, 0, arr.length - 1, tmp);
}
// Time: O(n*log(n)), Space: O(n)
public void sortIterative(int[] arr) {
if (arr == null || arr.length == 0) return;
int n = arr.length;
int[] tmp = new int[n];
for (int len = 1; len < n; len = 2*len) {
for (int low = 0; low < n; low += 2*len) {
int mid = Math.min(low+len-1, n-1);
int high = Math.min(low+2*len-1, n-1);
merge(arr, low, mid, high, tmp);
}
}
}
}
堆排序
視頻鏈接:https://algocasts.io/series/sorting-algorithms/episodes/jwmBqnW8
截圖:
代碼:
public class HeapSort {
private void swap(int[] arr, int i, int j) {
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
}
// Time: O(log(n))
private void siftDown(int[] arr, int i, int end) {
int parent = i, child = 2 * parent + 1;
while (child <= end) {
if (child+1 <= end && arr[child+1] > arr[child]) ++child;
if (arr[parent] >= arr[child]) break;
swap(arr, parent, child);
parent = child;
child = 2 * parent + 1;
}
}
// i 從 end/2 開始即可,因爲在二叉堆中,更大的 i 是沒有子節點的,沒必要做 siftDown
// Time: O(n)
// Reference:
// * https://www.geeksforgeeks.org/time-complexity-of-building-a-heap/
// * https://www2.cs.sfu.ca/CourseCentral/307/petra/2009/SLN_2.pdf
private void buildMaxHeap(int[] arr, int end) {
for (int i = end/2; i >= 0; --i) {
siftDown(arr, i, end);
}
}
// Time: O(n*log(n)), Space: O(1)
public void sort(int[] arr) {
if (arr == null || arr.length == 0) return;
buildMaxHeap(arr, arr.length - 1);
for (int end = arr.length - 1; end > 0; --end) {
swap(arr, 0, end);
siftDown(arr, 0, end - 1);
}
}
}
計數排序
視頻鏈接:https://algocasts.io/series/sorting-algorithms/episodes/XOp19ap2
截圖:
代碼:
public class CountingSort {
// indexes 最後存儲的是排序後,相同數字在結果數組的開始位置,相同數字會依次向後(右)填充。
// Time: O(n+k), Space: O(n+k)
public void sortLeft2Right(int[] arr) {
if (arr == null || arr.length == 0) return;
int max = arr[0], min = arr[0];
for (int num: arr) {
if (num > max) max = num;
if (num < min) min = num;
}
int k = max - min;
int[] indexes = new int[k+1];
for (int num: arr) ++indexes[num-min];
int start = 0;
for (int i = 0; i <= k; ++i) {
int count = indexes[i];
indexes[i] = start;
start += count;
}
int[] tmp = new int[arr.length];
for (int num: arr) {
int idx = indexes[num-min];
tmp[idx] = num;
++indexes[num-min];
}
System.arraycopy(tmp, 0, arr, 0, arr.length);
}
// indexes 最後存儲的是排序後,相同數字在結果數組的結束位置,相同數字會依次向前(左)填充。
// Time: O(n+k), Space: O(n+k)
public void sortRight2Left(int[] arr) {
if (arr == null || arr.length == 0) return;
int max = arr[0], min = arr[0];
for (int num: arr) {
if (num > max) max = num;
if (num < min) min = num;
}
int k = max - min;
int[] indexes = new int[k+1];
for (int num: arr) ++indexes[num-min];
--indexes[0];
for (int i = 1; i <= k; ++i)
indexes[i] = indexes[i] + indexes[i-1];
int[] tmp = new int[arr.length];
for (int i = arr.length-1; i >= 0; --i) {
int idx = indexes[arr[i]-min];
tmp[idx] = arr[i];
--indexes[arr[i]-min];
}
System.arraycopy(tmp, 0, arr, 0, arr.length);
}
}
桶排序
視頻鏈接:https://algocasts.io/series/sorting-algorithms/episodes/VBpL2omD
截圖:
代碼:
public class BucketSort {
private void insertionSort(List<Integer> arr) {
if (arr == null || arr.size() == 0) return;
for (int i = 1; i < arr.size(); ++i) {
int cur = arr.get(i);
int j = i - 1;
while (j >= 0 && arr.get(j) > cur) {
arr.set(j+1, arr.get(j));
--j;
}
arr.set(j+1, cur);
}
}
// 每個桶的大小,由於桶內使用插入排序,因此桶的大小使用一個較小值會比較高效。
//
// 一般來說,當處理的數組大小在 5-15 時,使用插入排序往往會比快排或歸併更高效。
// 因此在桶排序中,我們儘量讓單個桶內的元素個數是在 5-15 個之間,這樣可以用插入排序高效地完成桶內排序。
// 參考鏈接:https://algs4.cs.princeton.edu/23quicksort/
// 參考段落:
// Cutoff to insertion sort. As with mergesort,
// it pays to switch to insertion sort for tiny arrays.
// The optimum value of the cutoff is system-dependent,
// but any value between 5 and 15 is likely to work well in most situations.
private int bucketSize;
public BucketSort(int bucketSize) {
this.bucketSize = bucketSize;
}
// Time(avg): O(n+k), Time(worst): O(n^2), Space: O(n)
public void sort(int[] arr) {
if (arr == null || arr.length == 0) return;
int max = arr[0], min = arr[0];
for (int num: arr) {
if (num > max) max = num;
if (num < min) min = num;
}
int bucketCount = arr.length / bucketSize;
List<List<Integer>> buckets = new ArrayList<>(bucketCount);
for (int i = 0; i < bucketCount; ++i)
buckets.add(new ArrayList<>());
for (int num: arr) {
int idx = (int)((num - min) / (max - min + 1.0) * bucketCount);
buckets.get(idx).add(num);
}
int idx = 0;
for (List<Integer> bucket: buckets) {
insertionSort(bucket);
for (int num: bucket)
arr[idx++] = num;
}
}
}
基數排序
視頻鏈接:https://algocasts.io/series/sorting-algorithms/episodes/q2m595mz
截圖:
代碼:
public class RadixSort {
/**
* @param arr 待排數組
* @param bits 每次處理的二進制位數(可選值:1, 2, 4, 8, 16)
* @param mask 每次移動 bits 個二進制位後,使用 mask 取出最低的 bits 位。
*/
// b 表示每次處理的二進制位數
// Time: O(32/b * n), Space: O(n + 2^b)
private void sort(int[] arr, int bits, int mask) {
if (arr == null || arr.length == 0) return;
int n = arr.length, cnt = 32/bits;
int[] tmp = new int[n];
int[] indexes = new int[1<<bits];
for (int d = 0; d < cnt; ++d) {
for (int num: arr) {
int idx = (num >> (bits*d)) & mask;
++indexes[idx];
}
--indexes[0];
for (int i = 1; i < indexes.length; ++i)
indexes[i] = indexes[i] + indexes[i-1];
for (int i = n-1; i >= 0; --i) {
int idx = (arr[i] >> (bits*d)) & mask;
tmp[indexes[idx]] = arr[i];
--indexes[idx];
}
Arrays.fill(indexes, 0);
int[] t = arr;
arr = tmp;
tmp = t;
}
// handle the negative number
// get the length of positive part
int len = 0;
for (; len < n; ++len)
if (arr[len] < 0) break;
System.arraycopy(arr, len, tmp, 0, n-len); // copy negative part to tmp
System.arraycopy(arr, 0, tmp, n-len, len); // copy positive part to tmp
System.arraycopy(tmp, 0, arr, 0, n); // copy back to arr
}
// 基數爲 256,每次取 8 個二進制位作爲一個部分進行處理,32 位整數需要處理 4 次。
// 每次取出的 8 個二進制位會作爲計數排序的鍵值,去排序原始數據。
// 每次處理 8 個二進制位,是時間/空間上比較折衷的方法。
// 如果一次處理 16 個二進制位,速度會稍微快一些。但需要額外的空間是 2^16 = 65536,遠大於每次處理 8 個二進制位所需空間。
// 如果一次只處理 4 個二進制位,速度則會慢很多。
public void sort4pass(int[] arr) {
sort(arr, 8, 0xff);
}
// 基數爲 16,每次取 4 個二進制位作爲一個部分進行處理。32 位整數需要處理 8 次。
// 時間上比起 sort4pass 要差很多。
public void sort8pass(int[] arr) {
sort(arr, 4, 0x0f);
}
// 基數爲 65536,每次取 16 個二進制位作爲一個部分進行處理。32 位整數需要處理 2 次。
// 時間上比 sort4pass 要稍微好一些,但額外要使用多得多的空間。
public void sort2pass(int[] arr) {
sort(arr, 16, 0xffff);
}
// 基數爲 2,每次取 1 個二進制位作爲一個部分進行處理。32 位整數需要處理 32 次。
// 時間上比快排要差很多。
public void sort32pass(int[] arr) {
sort(arr, 1, 1);
}
// 基數爲 4,每次取 2 個二進制位作爲一個部分進行處理。32 位整數需要處理 16 次。
// 我是打醬油的。
public void sort16pass(int[] arr) {
sort(arr, 2, 3);
}
}