给定一个由 N 个整数组成的数组 arr[]和一个整数 K ，任务是根据它们与给定整数K的距离对这些整数进行排序。如果超过1 个元素位于相同的距离，则按升序打印它们。
注意：数组中两个元素之间的距离是通过它们的索引之间的差异来衡量的。
注意：整数 K 始终存在于数组arr[] 中并且是唯一的。
例子：
Input: arr[] = {12, 10, 102, 31, 15}, K = 102
Output: 102 10 31 12 15
Explanation:
Elements at their respective distance from K are,
At distance 0: 102
At distance 1: 10, 31 in sorted form.
At distance 2: 12, 15 in sorted form.
Hence, our resultant array is [ 102, 10, 31, 12, 15 ]
Input: arr[] = {14, 1101, 10, 35, 0}, K = 35
Output: 35 0 10 1101 14
Explanation:
Elements at their respective distance from K are,
At distance 0: 35
At distance 1: 10, 0 and in sorted form we have 0, 10.
At distance 2: 1101
At distance 3: 14
Hence, our resultant array is [ 35, 0, 10, 1101, 14 ]
方法 ：
为了解决上面提到的问题，我们创建了一个辅助向量来存储距离 K 任意距离的元素。然后找到给定整数 K 在数组 arr[] 中的位置，并将元素 K 插入到辅助向量中的位置 0 处。从 K 向左遍历数组，并将这些元素插入向量中与 K 的距离处。对 K 的右侧元素重复上述过程。最后，按排序顺序打印距离为 0 的数组元素。
下面是上述方法的实现：
C++
// C++ implementation to Sort the integers in
// array according to their distance from given
// element K present in the array
#include
using namespace std;
// Function to get sorted array based on
// their distance from given integer K
void distanceSort(int arr[], int K, int n)
{
// Vector to store respective elements
// with their distance from integer K
vector vd[n];
// Find the position of integer K
int pos;
for (int i = 0; i < n; i++) {
if (arr[i] == K) {
pos = i;
break;
}
}
// Insert the elements with their
// distance from K in vector
int i = pos - 1, j = pos + 1;
// Element at distnce 0
vd[0].push_back(arr[pos]);
// Elements at left side of K
while (i >= 0) {
vd[pos - i].push_back(arr[i]);
--i;
}
// Elements at right side of K
while (j < n) {
vd[j - pos].push_back(arr[j]);
++j;
}
// Print the vector content in sorted order
for (int i = 0; i <= max(pos, n - pos - 1); ++i) {
// Sort elements at same distance
sort(begin(vd[i]), end(vd[i]));
// Print elements at distance i from K
for (auto element : vd[i])
cout << element << " ";
}
}
// Driver code
int main()
{
int arr[] = {14, 1101, 10, 35, 0 }, K = 35;
int n = sizeof(arr) / sizeof(arr[0]);
distanceSort(arr, K, n);
return 0;
} Java
// Java implementation to Sort the integers in
// array according to their distance from given
// element K present in the array
import java.util.*;
class GFG{
// Function to get sorted array based on
// their distance from given integer K
@SuppressWarnings("unchecked")
static void distanceSort(int arr[], int K, int n)
{
// Vector to store respective elements
// with their distance from integer K
Vector vd[] = new Vector[n];
for(int i = 0; i < n; i++)
{
vd[i] = new Vector();
}
// Find the position of integer K
int pos = 0;
for(int i = 0; i < n; i++)
{
if (arr[i] == K)
{
pos = i;
break;
}
}
// Insert the elements with their
// distance from K in vector
int i = pos - 1, j = pos + 1;
// Element at distnce 0
vd[0].add(arr[pos]);
// Elements at left side of K
while (i >= 0)
{
vd[pos - i].add(arr[i]);
--i;
}
// Elements at right side of K
while (j < n)
{
vd[j - pos].add(arr[j]);
++j;
}
// Print the vector content in sorted order
for(i = 0; i <= Math.max(pos, n - pos - 1); ++i)
{
// Sort elements at same distance
Collections.sort(vd[i]);
// Print elements at distance i from K
for (j = 0; j < vd[i].size(); j++)
{
int element = (int)vd[i].get(j);
System.out.print(element + " ");
}
}
}
// Driver Code
public static void main(String s[])
{
int arr[] = {14, 1101, 10, 35, 0 };
int K = 35;
int n = arr.length;
distanceSort(arr, K, n);
}
}
// This code is contributed by rutvik_56Python3
# Python3 implementation to Sort the integers in
# array according to their distance from given
# element K present in the array
# Function to get sorted array based on
# their distance from given integer K
def distanceSort(arr,K,n):
# Vector to store respective elements
# with their distance from integer K
vd = [[] for i in range(n)]
# Find the position of integer K
for i in range(n):
if (arr[i] == K):
pos = i
break
# Insert the elements with their
# distance from K in vector
i = pos - 1
j = pos + 1
# Element at distnce 0
vd[0].append(arr[pos])
# Elements at left side of K
while (i >= 0):
vd[pos - i].append(arr[i])
i -= 1
# Elements at right side of K
while (j < n):
vd[j - pos].append(arr[j])
j += 1
# Print the vector content in sorted order
for i in range(max(pos, n - pos - 1) + 1):
# Sort elements at same distance
vd[i].sort(reverse=False)
# Print elements at distance i from K
for element in vd[i]:
print(element,end = " ")
# Driver code
if __name__ == '__main__':
arr = [14, 1101, 10, 35, 0]
K = 35
n = len(arr)
distanceSort(arr, K, n)
# This code is contributed by Surendra_GangwarC#
// C# implementation to Sort the integers in
// array according to their distance from given
// element K present in the array
using System;
using System.Collections.Generic;
class GFG{
// Function to get sorted array based on
// their distance from given integer K
static void distanceSort(int []arr,
int K, int n)
{
// List to store respective elements
// with their distance from integer K
List []vd = new List[n];
int i ;
for(i = 0; i < n; i++)
{
vd[i] = new List();
}
// Find the position of integer K
int pos = 0;
for(i = 0; i < n; i++)
{
if (arr[i] == K)
{
pos = i;
break;
}
}
// Insert the elements with their
// distance from K in vector
int j = pos + 1;
i = pos - 1;
// Element at distnce 0
vd[0].Add(arr[pos]);
// Elements at left side of K
while (i >= 0)
{
vd[pos - i].Add(arr[i]);
--i;
}
// Elements at right side of K
while (j < n)
{
vd[j - pos].Add(arr[j]);
++j;
}
// Print the vector content in sorted order
for(i = 0; i <= Math.Max(pos,
n - pos - 1); ++i)
{
// Sort elements at same distance
vd[i].Sort();
// Print elements at distance i from K
for (j = 0; j < vd[i].Count; j++)
{
int element = (int)vd[i][j];
Console.Write(element + " ");
}
}
}
// Driver Code
public static void Main(String []args)
{
int []arr = {14, 1101, 10, 35, 0};
int K = 35;
int n = arr.Length;
distanceSort(arr, K, n);
}
}
// This code is contributed by shikhasingrajput 输出：
35 0 10 1101 14
时间复杂度： O(N)
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