将数组压缩成范围

给定大小为N的整数数组，任务是将连续整数打印为范围。

例子：

Input : N = 7, arr=[7, 8, 9, 15, 16, 20, 25]
Output : 7-9 15-16 20 25
Consecutive elements present are[ {7, 8, 9}, {15, 16}, {20}, {25} ]
Hence output the result as 7-9 15-16 20 25

Input : N = 6, arr=[1, 2, 3, 4, 5, 6]
Output : 1-6

方法：
该问题可以很容易地可视化为游程长度编码问题的一种变体。

• 首先对数组进行排序。
• 然后，启动一个while循环以遍历数组以检查连续的元素。在任何特定情况下，连续数字的结尾将由j-1表示，并由i开头。
• 如果i不落入循环中，则将i递增1，否则将j + 1递增，以使其跳至当前范围之外的下一个ith元素。

下面是上述方法的实现：

C++

// C++ program to compress the array ranges
#include 
using namespace std;
  
// Function to compress the array ranges
void compressArr(int arr[], int n)
{
    int i = 0, j = 0;
    sort(arr, arr + n);
    while (i < n) {
  
        // start iteration from the
        // ith array element
        j = i;
  
        // loop until arr[i+1] == arr[i]
        // and increment j
        while ((j + 1 < n) &&
                 (arr[j + 1] == arr[j] + 1)) {
            j++;
        }
  
        // if the program do not enter into
        // the above while loop this means that
        // (i+1)th element is not consecutive
        // to i th element
        if (i == j) {
            cout << arr[i] << " ";
  
            // increment i for next iteration
            i++;
        }
        else {
            // print the consecutive range found
            cout << arr[i] << "-" << arr[j] << " ";
  
            // move i jump directly to j+1
            i = j + 1;
        }
    }
}
  
// Driver code
int main()
{
  
    int n = 7;
    int arr[n] = { 1, 3, 4, 5, 6, 9, 10 };
  
    compressArr(arr, n);
}

Java

// Java program to compress the array ranges
import java.util.Arrays; 
  
class GFG
{
  
// Function to compress the array ranges
static void compressArr(int arr[], int n)
{
    int i = 0, j = 0;
    Arrays.sort(arr);
    while (i < n) 
    {
  
        // start iteration from the
        // ith array element
        j = i;
  
        // loop until arr[i+1] == arr[i]
        // and increment j
        while ((j + 1 < n) &&
                (arr[j + 1] == arr[j] + 1)) 
        {
            j++;
        }
  
        // if the program do not enter into
        // the above while loop this means that
        // (i+1)th element is not consecutive
        // to i th element
        if (i == j)
        {
            System.out.print( arr[i] + " ");
  
            // increment i for next iteration
            i++;
        }
        else 
        {
            // print the consecutive range found
            System.out.print( arr[i] + "-" + arr[j] + " ");
  
            // move i jump directly to j+1
            i = j + 1;
        }
    }
}
  
    // Driver code
    public static void main (String[] args)
    {
        int n = 7;
        int arr[] = { 1, 3, 4, 5, 6, 9, 10 };
  
        compressArr(arr, n);
    }
}
  
// This code is contributed by anuj_67..

Python3

# Python program to compress the array ranges
  
# Function to compress the array ranges
def compressArr(arr, n):
    i = 0;
    j = 0;
    arr.sort();
    while (i < n):
  
        # start iteration from the
        # ith array element
        j = i;
  
        # loop until arr[i+1] == arr[i]
        # and increment j
        while ((j + 1 < n) and
                (arr[j + 1] == arr[j] + 1)):
            j += 1;
  
        # if the program do not enter into
        # the above while loop this means that
        # (i+1)th element is not consecutive
        # to i th element
        if (i == j):
            print(arr[i], end=" ");
  
            # increment i for next iteration
            i+=1;
        else:
            # print the consecutive range found
            print(arr[i], "-", arr[j], end=" ");
  
            # move i jump directly to j+1
            i = j + 1;
  
# Driver code
n = 7;
arr = [ 1, 3, 4, 5, 6, 9, 10 ];
compressArr(arr, n);
  
# This code is contributed by PrinciRaj1992

C#

// C# program to compress the array ranges
using System;
  
class GFG
{
  
// Function to compress the array ranges
static void compressArr(int []arr, int n)
{
    int i = 0, j = 0;
    Array.Sort(arr);
    while (i < n) 
    {
  
        // start iteration from the
        // ith array element
        j = i;
  
        // loop until arr[i+1] == arr[i]
        // and increment j
        while ((j + 1 < n) &&
                (arr[j + 1] == arr[j] + 1)) 
        {
            j++;
        }
  
        // if the program do not enter into
        // the above while loop this means that
        // (i+1)th element is not consecutive
        // to i th element
        if (i == j)
        {
            Console.Write( arr[i] + " ");
  
            // increment i for next iteration
            i++;
        }
        else
        {
            // print the consecutive range found
            Console.Write( arr[i] + "-" + arr[j] + " ");
  
            // move i jump directly to j+1
            i = j + 1;
        }
    }
}
  
// Driver code
public static void Main ()
{
    int n = 7;
    int []arr = { 1, 3, 4, 5, 6, 9, 10 };
  
    compressArr(arr, n);
}
}
  
// This code is contributed by anuj_67..

输出：

1 3-6 9-10

时间复杂度： O(n)