给定一个由N 个自然数组成的未排序数组arr[]和缺失数为0使得arr[i] ≠ i ，任务是在不改变初始顺序的情况下找到并填充这些缺失数。请注意，数组只能包含一次从 1 到 N 的数字。
例子：
Input: arr[] = {7, 4, 0, 3, 0, 5, 1}
Output: 7 4 6 3 2 5 1
Explanation:
In the given array, unfilled indices are 2 and 4. So the missing numbers that can be filled, to fulfil the criteria arr[i] ≠ i, are 6 and 2 respectively.
Input: arr[] = {2, 1, 0, 0, 0}
Output: 2 1 4 5 3
Explanation:
In the given array unfilled indices are 3, 4 and 5. So the missing numbers that can be filled, to fulfil the criteria arr[i] ≠ i, are 4, 5 and 3 respectively.
方法：
- 将所有未填充的索引存储在数组unfilled_indices[] 中。即，对于所有i使得arr[i] = 0。
- 还要存储在数组missing[]中缺失的所有元素。这可以通过首先插入从1到N 的所有元素然后删除所有不为0 的元素来完成
- 简单地迭代unfilled_indices[]数组并开始分配存储在missing[]数组中的所有元素，可能从后向获取每个元素(以避免任何i得到arr[i] = i )。
- 现在有可能最多一个索引可以得到相同的arr[i] = i 。在检查其他索引是否应存在于unfilled_indices[]之后，只需与它交换任何其他元素。
- 打印新数组。
下面是上述方法的实现：
C++
// C++ implementation of above approach
#include
using namespace std;
// Function to print array
void printArray(int[], int);
// Function to fill the position
// with arr[i] = 0
void solve(int arr[], int n)
{
set unfilled_indices;
set missing;
// Inserting all elements in
// missing[] set from 1 to N
for (int i = 1; i < n; i++)
missing.insert(i);
for (int i = 1; i < n; i++) {
// Inserting unfilled positions
if (arr[i] == 0)
unfilled_indices.insert(i);
// Removing allocated_elements
else {
auto it = missing.find(arr[i]);
missing.erase(it);
}
}
auto it2 = missing.end();
it2--;
// Loop for filling the positions
// with arr[i] != i
for (auto it = unfilled_indices.begin();
it != unfilled_indices.end();
it++, it2--) {
arr[*it] = *it2;
}
int pos = 0;
// Checking for any arr[i] = i
for (int i = 1; i < n; i++) {
if (arr[i] == i) {
pos = i;
}
}
int x;
// Finding the suitable position
// in the array to swap with found i
// for which arr[i] = i
if (pos != 0) {
for (int i = 1; i < n; i++) {
if (pos != i) {
// Checking if the position
// is present in unfilled_position
if (unfilled_indices.find(i)
!= unfilled_indices.end()) {
// Swapping arr[i] & arr[pos]
// (arr[pos] = pos)
x = arr[i];
arr[i] = pos;
arr[pos] = x;
break;
}
}
}
}
printArray(arr, n);
}
// Function to Print the array
void printArray(int arr[], int n)
{
for (int i = 1; i < n; i++)
cout << arr[i] << " ";
}
// Driver Code
int main()
{
int arr[] = { 0, 7, 4, 0,
3, 0, 5, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
solve(arr, n);
return 0;
} Java
// Java implementation of above approach
import java.io.*;
import java.util.*;
class GFG
{
// Function to fill the position
// with arr[i] = 0
static void solve(int arr[], int n)
{
Set unfilled_indices = new HashSet();
Set missing = new HashSet();
// Inserting all elements in
// missing[] set from 1 to N
for (int i = 1; i < n; i++)
{
missing.add(i);
}
for (int i = 1; i < n; i++)
{
// Inserting unfilled positions
if (arr[i] == 0)
{
unfilled_indices.add(i);
}
// Removing allocated_elements
else
{
missing.remove(arr[i]);
}
}
int[] mi = new int[missing.size()];
int l = 0;
for(int x:missing)
{
mi[l++] = x;
}
int m = missing.size();
// Loop for filling the positions
// with arr[i] != i
for(int it:unfilled_indices)
{
arr[it] = mi[m - 1];
m--;
}
int pos = 0;
// Checking for any arr[i] = i
for (int i = 1; i < n; i++)
{
if (arr[i] == i)
{
pos = i;
}
}
int x;
// Finding the suitable position
// in the array to swap with found i
// for which arr[i] = i
if (pos != 0)
{
for (int i = 1; i < n; i++)
{
if (pos != i)
{
// Checking if the position
// is present in unfilled_position
if(unfilled_indices.contains(i))
{
// Swapping arr[i] & arr[pos]
// (arr[pos] = pos)
x = arr[i];
arr[i] = pos;
arr[pos] = x;
break;
}
}
}
}
printArray(arr, n);
}
// Function to Print the array
static void printArray(int arr[], int n)
{
for (int i = 1; i < n; i++)
{
System.out.print(arr[i] + " ");
}
}
// Driver Code
public static void main (String[] args)
{
int[] arr = { 0, 7, 4, 0,3, 0, 5, 1 };
int n = arr.length;
solve(arr, n);
}
}
// This code is contributed by avanitrachhadiya2155 Python3
# Python3 implementation of above approach
# Function to fill the position
# with arr[i] = 0
def solve(arr, n):
unfilled_indices = {}
missing = {}
# Inserting all elements in
# missing[] set from 1 to N
for i in range(n):
missing[i] = 1
for i in range(1, n):
# Inserting unfilled positions
if (arr[i] == 0):
unfilled_indices[i] = 1
# Removing allocated_elements
else:
del missing[arr[i]]
it2 = list(missing.keys())
m = len(it2)
# Loop for filling the positions
# with arr[i] != i
for it in unfilled_indices:
arr[it] = it2[m - 1]
m -= 1
pos = 0
# Checking for any arr[i] = i
for i in range(1, n):
if (arr[i] == i):
pos = i
x = 0
# Finding the suitable position
# in the array to swap with found i
# for which arr[i] = i
if (pos != 0):
for i in range(1, n):
if (pos != i):
# Checking if the position
# is present in unfilled_position
if i in unfilled_indices:
# Swapping arr[i] & arr[pos]
# (arr[pos] = pos)
x = arr[i]
arr[i] = pos
arr[pos] = x
break
printArray(arr, n)
# Function to Print the array
def printArray(arr, n):
for i in range(1, n):
print(arr[i], end = " ")
# Driver Code
if __name__ == '__main__':
arr = [ 0, 7, 4, 0, 3, 0, 5, 1 ]
n = len(arr)
solve(arr, n)
# This code is contributed by mohit kumar 29C#
// C# implementation of above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to fill the position
// with arr[i] = 0
static void solve(int[] arr, int n)
{
HashSet unfilled_indices = new HashSet();
HashSet missing = new HashSet();
// Inserting all elements in
// missing[] set from 1 to N
for(int i = 1; i < n; i++)
{
missing.Add(i);
}
for(int i = 1; i < n; i++)
{
// Inserting unfilled positions
if (arr[i] == 0)
{
unfilled_indices.Add(i);
}
// Removing allocated_elements
else
{
missing.Remove(arr[i]);
}
}
int[] mi = new int[missing.Count];
int l = 0;
foreach(int x in missing)
{
mi[l++] = x;
}
int m = missing.Count;
// Loop for filling the positions
// with arr[i] != i
foreach(int it in unfilled_indices)
{
arr[it] = mi[m - 1];
m--;
}
int pos = 0;
// Checking for any arr[i] = i
for(int i = 1; i < n; i++)
{
if (arr[i] == i)
{
pos = i;
}
}
// Finding the suitable position
// in the array to swap with found i
// for which arr[i] = i
if (pos != 0)
{
for(int i = 1; i < n; i++)
{
if (pos != i)
{
// Checking if the position
// is present in unfilled_position
if (unfilled_indices.Contains(i))
{
// Swapping arr[i] & arr[pos]
// (arr[pos] = pos)
int x = arr[i];
arr[i] = pos;
arr[pos] = x;
break;
}
}
}
}
printArray(arr, n);
}
// Function to Print the array
static void printArray(int[] arr, int n)
{
for(int i = 1; i < n; i++)
{
Console.Write(arr[i] + " ");
}
}
// Driver Code
static public void Main()
{
int[] arr = { 0, 7, 4, 0, 3, 0, 5, 1 };
int n = arr.Length;
solve(arr, n);
}
}
// This code is contributed by rag2127 输出：
7 4 6 3 2 5 1如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live