使用Java进行数据排序的10个函数
1. 冒泡排序(Bubble Sort): 该算法使用嵌套循环将较大的元素逐步交换到数组的末尾。时间复杂度为O(n^2)。以下是其实现代码:
public static void bubbleSort(int[] arr) {
int n = arr.length;
for (int i = 0; i < n-1; i++) {
for (int j = 0; j < n-i-1; j++) {
if (arr[j] > arr[j+1]) {
int temp = arr[j];
arr[j] = arr[j+1];
arr[j+1] = temp;
}
}
}
}
2. 选择排序(Selection Sort): 该算法每次选择最小的元素并将其放置在已排序部分的末尾。时间复杂度为O(n^2)。以下是其实现代码:
public static void selectionSort(int[] arr) {
int n = arr.length;
for (int i = 0; i < n-1; i++) {
int minIndex = i;
for (int j = i+1; j < n; j++) {
if (arr[j] < arr[minIndex]) {
minIndex = j;
}
}
int temp = arr[minIndex];
arr[minIndex] = arr[i];
arr[i] = temp;
}
}
3. 插入排序(Insertion Sort): 该算法将每个元素插入到已排序序列的适当位置。时间复杂度为O(n^2)。以下是其实现代码:
public static void insertionSort(int[] arr) {
int n = arr.length;
for (int i = 1; i < n; i++) {
int key = arr[i];
int j = i-1;
while (j >= 0 && arr[j] > key) {
arr[j+1] = arr[j];
j = j-1;
}
arr[j+1] = key;
}
}
4. 快速排序(Quick Sort): 该算法使用分治法原则,通过选择一个基准元素将数组分成两个子数组,然后对子数组进行排序。时间复杂度平均为O(nlogn)。以下是其实现代码:
public static void quickSort(int[] arr, int low, int high) {
if (low < high) {
int pi = partition(arr, low, high);
quickSort(arr, low, pi-1);
quickSort(arr, pi+1, high);
}
}
public static int partition(int[] arr, int low, int high) {
int pivot = arr[high];
int i = (low-1);
for (int j = low; j < high; j++) {
if (arr[j] < pivot) {
i++;
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
int temp = arr[i+1];
arr[i+1] = arr[high];
arr[high] = temp;
return i+1;
}
5. 归并排序(Merge Sort): 该算法使用分治法原则,将数组分成两半,分别对其进行排序,然后合并两个已排序的子数组。时间复杂度为O(nlogn)。以下是其实现代码:
public static void mergeSort(int[] arr, int low, int high) {
if (low < high) {
int mid = (low+high)/2;
mergeSort(arr, low, mid);
mergeSort(arr, mid+1, high);
merge(arr, low, mid, high);
}
}
public static void merge(int[] arr, int low, int mid, int high) {
int n1 = mid - low + 1;
int n2 = high - mid;
int[] left = new int[n1];
int[] right = new int[n2];
for (int i = 0; i < n1; i++) {
left[i] = arr[low+i];
}
for (int j = 0; j < n2; j++) {
right[j] = arr[mid+1+j];
}
int i = 0, j = 0, k = low;
while (i < n1 && j < n2) {
if (left[i] <= right[j]) {
arr[k] = left[i];
i++;
} else {
arr[k] = right[j];
j++;
}
k++;
}
while (i < n1) {
arr[k] = left[i];
i++;
k++;
}
while (j < n2) {
arr[k] = right[j];
j++;
k++;
}
}
6. 堆排序(Heap Sort): 该算法使用二叉堆数据结构,通过反复将最大元素从堆顶移除,并将其放置在数组的末尾来排序。时间复杂度为O(nlogn)。以下是其实现代码:
public static void heapSort(int[] arr) {
int n = arr.length;
for (int i = n/2-1; i >= 0; i--) {
heapify(arr, n, i);
}
for (int i = n-1; i >= 0; i--) {
int temp = arr[0];
arr[0] = arr[i];
arr[i] = temp;
heapify(arr, i, 0);
}
}
public static void heapify(int[] arr, int n, int i) {
int largest = i;
int left = 2*i + 1;
int right = 2*i + 2;
if (left < n && arr[left] > arr[largest]) {
largest = left;
}
if (right < n && arr[right] > arr[largest]) {
largest = right;
}
if (largest != i) {
int temp = arr[i];
arr[i] = arr[largest];
arr[largest] = temp;
heapify(arr, n, largest);
}
}
7. 计数排序(Counting Sort): 该算法通过确定一个元素在整个序列中的位置来排序,适用于非负整数。时间复杂度为O(n+k),其中k是整数范围。以下是其实现代码:
public static void countingSort(int[] arr) {
int n = arr.length;
int max = Arrays.stream(arr).max().getAsInt();
int min = Arrays.stream(arr).min().getAsInt();
int range = max - min + 1;
int[] count = new int[range];
int[] output = new int[n];
for (int i = 0; i < n; i++) {
count[arr[i]-min]++;
}
for (int i = 1; i < range; i++) {
count[i] += count[i-1];
}
for (int i = n-1; i >= 0; i--) {
output[count[arr[i]-min]-1] = arr[i];
count[arr[i]-min]--;
}
for (int i = 0; i < n; i++) {
arr[i] = output[i];
}
}
8. 基数排序(Radix Sort): 该算法按照元素的位数进行排序,从最低有效位到最高有效位。时间复杂度为O(nk),其中k是元素的最大位数。以下是其实现代码:
`java
public static void radixSort(int[] arr) {
int max = Arrays.stream(arr).max().getAsInt();
for (int exp = 1; max/exp > 0; exp *= 10) {
countingSort(arr, exp);
}
}
public static void countingSort(int[] arr, int exp) {
int n = arr.length;
int[] output = new int[n];
int[] count = new int[10];
for (int i = 0; i < n; i++) {
count[(arr[i]/exp)%10]++;
}
for (int i = 1; i < 10; i++) {
count[i] += count[i-1];
}
for (int i = n-1; i >= 0; i--) {
output[count[(arr[i]/exp)%10]-1] = arr[i];
count[(arr[i]/exp)%10]--;
}
