给定具有n个不同元素的数组，请将给定数组转换为所有元素都在0到n-1范围内的形式。元素的顺序相同，即0代替最小元素，1代表第二个最小元素，…n-1代表最大元素。
Input: arr[] = {10, 40, 20}
Output: arr[] = {0, 2, 1}
Input: arr[] = {5, 10, 40, 30, 20}
Output: arr[] = {0, 1, 4, 3, 2}
预期的时间复杂度为O(n Log n)。
我们强烈建议您单击此处并进行实践，然后再继续解决方案。
方法1(简单)
一个简单的解决方案是首先找到最小元素，将其替换为0，考虑剩余数组，然后在剩余数组中找到最小值，然后将其替换为1，依此类推。该解决方案的时间复杂度为O(n 2 )
方法2(高效)
这个想法是使用哈希和排序。以下是步骤。
1)创建一个临时数组并将给定数组的内容复制到temp []。这需要O(n)时间。
2)按升序对temp []进行排序。这需要O(n Log n)时间。
3)创建一个空的哈希表。这需要O(1)时间。
4)从左到右遍历temp []形式，并将数字及其值的映射(在转换后的数组中)存储在哈希表中。这平均需要O(n)时间。
5)遍历给定的数组，并使用哈希表将元素更改为其位置。这平均需要O(n)时间。
该解决方案的总时间复杂度为O(n Log n)。
以下是上述想法的实现。
C++
// C++ program to convert an array in reduced
// form
#include
using namespace std;
void convert(int arr[], int n)
{
// Create a temp array and copy contents
// of arr[] to temp
int temp[n];
memcpy(temp, arr, n*sizeof(int));
// Sort temp array
sort(temp, temp + n);
// Create a hash table. Refer
// http://tinyurl.com/zp5wgef
unordered_map umap;
// One by one insert elements of sorted
// temp[] and assign them values from 0
// to n-1
int val = 0;
for (int i = 0; i < n; i++)
umap[temp[i]] = val++;
// Convert array by taking positions from
// umap
for (int i = 0; i < n; i++)
arr[i] = umap[arr[i]];
}
void printArr(int arr[], int n)
{
for (int i=0; i Java
// Java Program to convert an Array
// to reduced form
import java.util.*;
class GFG
{
public static void convert(int arr[], int n)
{
// Create a temp array and copy contents
// of arr[] to temp
int temp[] = arr.clone();
// Sort temp array
Arrays.sort(temp);
// Create a hash table.
HashMap umap = new HashMap<>();
// One by one insert elements of sorted
// temp[] and assign them values from 0
// to n-1
int val = 0;
for (int i = 0; i < n; i++)
umap.put(temp[i], val++);
// Convert array by taking positions from
// umap
for (int i = 0; i < n; i++)
arr[i] = umap.get(arr[i]);
}
public static void printArr(int arr[], int n)
{
for (int i = 0; i < n; i++)
System.out.print(arr[i] + " ");
}
// Driver code
public static void main(String[] args)
{
int arr[] = {10, 20, 15, 12, 11, 50};
int n = arr.length;
System.out.println("Given Array is ");
printArr(arr, n);
convert(arr , n);
System.out.println("\n\nConverted Array is ");
printArr(arr, n);
}
}
// This code is contributed by Abhishek Panwar Python3
# Python3 program to convert an array
# in reduced form
def convert(arr, n):
# Create a temp array and copy contents
# of arr[] to temp
temp = [arr[i] for i in range (n) ]
# Sort temp array
temp.sort()
# create a map
umap = {}
# One by one insert elements of sorted
# temp[] and assign them values from 0
# to n-1
val = 0
for i in range (n):
umap[temp[i]] = val
val += 1
# Convert array by taking positions from umap
for i in range (n):
arr[i] = umap[arr[i]]
def printArr(arr, n):
for i in range(n):
print(arr[i], end = " ")
# Driver Code
if __name__ == "__main__":
arr = [10, 20, 15, 12, 11, 50]
n = len(arr)
print("Given Array is ")
printArr(arr, n)
convert(arr , n)
print("\n\nConverted Array is ")
printArr(arr, n)
# This code is contributed by Abhishek GuptaC#
// C# Program to convert an Array
// to reduced form
using System;
using System.Collections.Generic;
using System.Linq;
class GFG
{
public static void convert(int []arr, int n)
{
// Create a temp array and copy contents
// of []arr to temp
int []temp = new int[arr.Length];
Array.Copy(arr, 0, temp, 0, arr.Length);
// Sort temp array
Array.Sort(temp);
// Create a hash table.
Dictionary umap =
new Dictionary();
// One by one insert elements of sorted
// []temp and assign them values from 0
// to n - 1
int val = 0;
for (int i = 0; i < n; i++)
if(umap.ContainsKey(temp[i]))
umap[temp[i]] = val++;
else
umap.Add(temp[i], val++);
// Convert array by taking positions from
// umap
for (int i = 0; i < n; i++)
arr[i] = umap[arr[i]];
}
public static void printArr(int []arr, int n)
{
for (int i = 0; i < n; i++)
Console.Write(arr[i] + " ");
}
// Driver code
public static void Main(String[] args)
{
int []arr = {10, 20, 15, 12, 11, 50};
int n = arr.Length;
Console.WriteLine("Given Array is ");
printArr(arr, n);
convert(arr , n);
Console.WriteLine("\n\nConverted Array is ");
printArr(arr, n);
}
}
// This code is contributed by PrinciRaj1992 输出 ：
Given Array is
10 20 15 12 11 50
Converted Array is
0 4 3 2 1 5
将数组转换为简化形式|集合2(使用向量对)