对近乎排序(或 K 排序)的数组进行排序 | Set 2 (Gap 方法 – Shell 排序)
给定一个包含N个元素的数组arr[] ,其中每个元素距其目标位置最多K ,任务是设计一种在O(N*log(K))时间内排序的算法。
例子:
Input: arr[] = {10, 9, 8, 7, 4, 70, 60, 50}, K = 4
Output: 4 7 8 9 10 50 60 70
Explanation:
Follow the steps below to sort the array:
- Start with Gap = K(i.e. 4)
- 10 9 8 7 4 70 60 50, swap the elements at indices 0 and 4. Then the array modifies to {4, 9, 8, 7, 10, 70, 60, 50}.
4 9 8 7 10 70 60 50, Do not swap the elements at indices 1 and 5.
4 9 8 7 10 70 60 50, Do not swap the elements at indices 2 and 6.
4 9 8 7 10 70 60 50, Do not swap the elements at indices 3 and 7.
- 10 9 8 7 4 70 60 50, swap the elements at indices 0 and 4. Then the array modifies to {4, 9, 8, 7, 10, 70, 60, 50}.
- Gap = ceiling of 4/2 = 2
- 4 9 8 7 10 70 60 50, Do not swap the elements at indices 0 and 2.
4 9 8 7 10 70 60 50, swap the elements at indices 1 and 3. Then the array modifies to {4, 7, 8, 9, 10, 70, 60, 50}.
4 7 8 9 10 70 60 50, Do not swap the elements at indices 2 and 4.
4 7 8 9 10 70 60 50, Do not swap the elements at indices 3 and 5.
4 7 8 9 10 70 60 50, Do not swap the elements at indices 4 and 6.
4 7 8 9 10 70 60 50, swap the elements at indices 5 and 7. Then the array modifies to {4, 7, 8, 9, 10, 70, 60, 50}.
4 7 8 9 10 50 60 70
- 4 9 8 7 10 70 60 50, Do not swap the elements at indices 0 and 2.
- Gap = ceiling of 2/2 = 1
- 4 7 8 9 10 50 60 70, Do not swap the elements at indices 0 and 1.
4 7 8 9 10 50 60 70, Do not swap the elements at indices 1 and 2.
4 7 8 9 10 50 60 70, Do not swap the elements at indices 2 and 3.
4 7 8 9 10 50 60 70, Do not swap the elements at indices 3 and 4.
4 7 8 9 10 50 60 70, Do not swap the elements at indices 4 and 5.
4 7 8 9 10 50 60 70, Do not swap the elements at indices 5 and 6.
4 7 8 9 10 50 60 70, Do not swap the elements at indices 6 and 7.
- 4 7 8 9 10 50 60 70, Do not swap the elements at indices 0 and 1.
Input: arr[] = {6, 5, 3, 2, 8, 10, 9}, K = 3
Output: 2 3 5 6 8 9 10
方法:给定的问题对几乎排序(或 K 排序)的数组进行排序已经解决。这里的想法是使用 shell 排序对数组进行排序。这里使用的想法类似于 In-Place Merge Sort 的合并步骤。请按照以下步骤解决问题:
- 初始化一个变量,例如Gap ,其值为K以对每个Gap th进行排序 每个子列表的元素。
- 迭代直到Gap大于0并执行以下步骤:
- 使用变量i在范围[0, N-Gap]上进行迭代,并且在每次迭代中,如果arr[i]大于arr[i+Gap],则交换数组元素。
- 将Gap更新为Gap = ceil(Gap/2)。
- 最后,完成上述步骤后,打印数组arr[]的元素。
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the nextGap
int nextGap(double k)
{
if (k < 2)
return 0;
return ceil(k / 2);
}
// A utility function to print the array
void printArray(int arr[], int n)
{
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
}
// Function to sort a K sorted array
void kSort(int arr[], int K, int n)
{
// Iterate until gap is atleast
// greater than 0
for (int gap = K; gap > 0; gap = nextGap(gap)) {
// Iterate over the range [0, N]
for (int i = 0; i + gap < n; i++) {
// If arr[i] is greater
// than arr[i+gap]
if (arr[i] > arr[i + gap]) {
// Swap arr[i] and
// arr[i+gap]
swap(arr[i], arr[i + gap]);
}
}
}
printArray(arr, n);
}
// Driver Code
int main()
{
// Input
int arr[] = { 10, 9, 8, 7, 4, 70, 60, 50 };
int K = 3;
int n = sizeof(arr) / sizeof(arr[0]);
// Function call
kSort(arr, K, n);
return 0;
}
// This code is contributed by lokesh potta.
Java
// Java program for the above approach
import java.util.Iterator;
import java.util.PriorityQueue;
class GFG {
// Function to sort a K sorted array
static void kSort(int[] arr, int K)
{
// Iterate until gap is atleast
// greater than 0
for (int gap = K; gap > 0; gap = nextGap(gap)) {
// Iterate over the range [0, N]
for (int i = 0; i + gap < arr.length; i++) {
// If arr[i] is greater
// than arr[i+gap]
if (arr[i] > arr[i + gap]) {
// Swap arr[i] and
// arr[i+gap]
swap(arr, i, i + gap);
}
}
}
printArray(arr);
}
// Function to find the nextGap
static int nextGap(double k)
{
if (k < 2)
return 0;
return (int)Math.ceil(k / 2);
}
// Function to swap two elements
// of the array arr[]
static void swap(int[] arr, int i, int j)
{
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
// A utility function to print the array
private static void printArray(int[] arr)
{
for (int i = 0; i < arr.length; i++)
System.out.print(arr[i] + " ");
}
// Driver Code
public static void main(String[] args)
{
// Input
int arr[] = { 10, 9, 8, 7, 4, 70, 60, 50 };
int K = 3;
// Function call
kSort(arr, K);
}
}
Python3
# Python3 program for the above approach
import math
# Function to find the nextGap
def nextGap(k):
if (k < 2):
return 0
return math.ceil(k / 2)
# A utility function to print array
def printArray(arr, n):
for i in range(n):
print(arr[i], end = " ")
# Function to sort a K sorted array
def kSort(arr, K, n):
# Iterate until gap is atleast
# greater than 0
gap = K
while (gap > 0):
# Iterate over the range [0, N]
i = 0
while (i + gap < n):
# If arr[i] is greater
# than arr[i+gap]
if (arr[i] > arr[i + gap]):
# Swap arr[i] and
# arr[i+gap]
arr[i], arr[i + gap] = arr[i + gap], arr[i]
i += 1
gap = nextGap(gap)
printArray(arr, n)
# Driver Code
# Input
arr = [ 10, 9, 8, 7, 4, 70, 60, 50 ]
K = 3
n = len(arr)
# Function call
kSort(arr, K, n)
# This code is contributed by target_2
C#
// C# program for the above approach
using System;
class GFG {
// Function to sort a K sorted array
static void kSort(int[] arr, int K)
{
// Iterate until gap is atleast
// greater than 0
for (int gap = K; gap > 0; gap = nextGap(gap)) {
// Iterate over the range [0, N]
for (int i = 0; i + gap < arr.Length; i++) {
// If arr[i] is greater
// than arr[i+gap]
if (arr[i] > arr[i + gap]) {
// Swap arr[i] and
// arr[i+gap]
swap(arr, i, i + gap);
}
}
}
printArray(arr);
}
// Function to find the nextGap
static int nextGap(double k)
{
if (k < 2)
return 0;
return (int)Math.Ceiling(k / 2);
}
// Function to swap two elements
// of the array arr[]
static void swap(int[] arr, int i, int j)
{
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
// A utility function to print the array
private static void printArray(int[] arr)
{
for (int i = 0; i < arr.Length; i++)
Console.Write(arr[i] + " ");
}
// Driver Code
public static void Main(string[] args)
{
// Input
int []arr = { 10, 9, 8, 7, 4, 70, 60, 50 };
int K = 3;
// Function call
kSort(arr, K);
}
}
// This code is contributed by ukasp.
Javascript
输出
4 7 8 9 10 50 60 70
时间复杂度: O(N*log K)
辅助空间: O(1)