给定一个包含 n 个元素的数组 A。我们需要使用数组中的最小替换将数组更改为从 1 到 n 的数字排列。
例子：
Input : A[] = {2, 2, 3, 3}
Output : 2 1 3 4
Explanation:
To make it a permutation of 1 to 4, 1 and 4 are
missing from the array. So replace 2, 3 with
1 and 4.
Input : A[] = {1, 3, 2}
Output : 1 3 2
Input : A[] = {10, 1, 2}
Output : 3 1 2
方法：请注意，我们不需要更改 [1, n] 范围内且不同的数字（仅出现一次）。所以，我们使用贪婪的方法。如果我们遇到了以前从未遇到过的数字，并且这个数字在 1 和 n 之间，我们就保持这个数字不变。并删除重复元素并添加范围 [1, n] 中的缺失元素。还要替换数字，而不是在范围内。
C++
// CPP program to make a permutation of numbers
// from 1 to n using minimum changes.
#include
using namespace std;
void makePermutation(int a[], int n)
{
// Store counts of all elements.
unordered_map count;
for (int i = 0; i < n; i++)
count[a[i]]++;
int next_missing = 1;
for (int i = 0; i < n; i++) {
if (count[a[i]] != 1 || a[i] > n || a[i] < 1) {
count[a[i]]--;
// Find next missing element to put
// in place of current element.
while (count.find(next_missing) != count.end())
next_missing++;
// Replace with next missing and insert the
// missing element in hash.
a[i] = next_missing;
count[next_missing] = 1;
}
}
}
// Driver Code
int main()
{
int A[] = { 2, 2, 3, 3 };
int n = sizeof(A) / sizeof(A[0]);
makePermutation(A, n);
for (int i = 0; i < n; i++)
cout << A[i] << " ";
return 0;
} Java
// Java program to make a permutation of numbers
// from 1 to n using minimum changes.
import java.util.*;
class GFG
{
static void makePermutation(int []a, int n)
{
// Store counts of all elements.
HashMap count = new HashMap();
for (int i = 0; i < n; i++)
{
if(count.containsKey(a[i]))
{
count.put(a[i], count.get(a[i]) + 1);
}
else
{
count.put(a[i], 1);
}
}
int next_missing = 1;
for (int i = 0; i < n; i++)
{
if (count.containsKey(a[i]) &&
count.get(a[i]) != 1 ||
a[i] > n || a[i] < 1)
{
count.put(a[i], count.get(a[i]) - 1);
// Find next missing element to put
// in place of current element.
while (count.containsKey(next_missing))
next_missing++;
// Replace with next missing and insert
// the missing element in hash.
a[i] = next_missing;
count. put(next_missing, 1);
}
}
}
// Driver Code
public static void main(String[] args)
{
int A[] = { 2, 2, 3, 3 };
int n = A.length;
makePermutation(A, n);
for (int i = 0; i < n; i++)
System.out.print(A[i] + " ");
}
}
// This code is contributed by PrinciRaj1992 Python3
# Python3 code to make a permutation
# of numbers from 1 to n using
# minimum changes.
def makePermutation (a, n):
# Store counts of all elements.
count = dict()
for i in range(n):
if count.get(a[i]):
count[a[i]] += 1
else:
count[a[i]] = 1;
next_missing = 1
for i in range(n):
if count[a[i]] != 1 or a[i] > n or a[i] < 1:
count[a[i]] -= 1
# Find next missing element to put
# in place of current element.
while count.get(next_missing):
next_missing+=1
# Replace with next missing and
# insert the missing element in hash.
a[i] = next_missing
count[next_missing] = 1
# Driver Code
A = [ 2, 2, 3, 3 ]
n = len(A)
makePermutation(A, n)
for i in range(n):
print(A[i], end = " ")
# This code is contributed by "Sharad_Bhardwaj".C#
// C# program to make a permutation of numbers
// from 1 to n using minimum changes.
using System;
using System.Collections.Generic;
class GFG
{
static void makePermutation(int []a, int n)
{
// Store counts of all elements.
Dictionary count = new Dictionary();
for (int i = 0; i < n; i++)
{
if(count.ContainsKey(a[i]))
{
count[a[i]] = count[a[i]] + 1;
}
else
{
count.Add(a[i], 1);
}
}
int next_missing = 1;
for (int i = 0; i < n; i++)
{
if (count.ContainsKey(a[i]) &&
count[a[i]] != 1 ||
a[i] > n || a[i] < 1)
{
count[a[i]] = count[a[i]] - 1;
// Find next missing element to put
// in place of current element.
while (count.ContainsKey(next_missing))
next_missing++;
// Replace with next missing and insert
// the missing element in hash.
a[i] = next_missing;
count.Add(next_missing, 1);
}
}
}
// Driver Code
public static void Main(String[] args)
{
int []A = { 2, 2, 3, 3 };
int n = A.Length;
makePermutation(A, n);
for (int i = 0; i < n; i++)
Console.Write(A[i] + " ");
}
}
// This code is contributed by Princi Singh Javascript
输出：
1 2 4 3如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。