在给定N – 1的情况下，包含N个数字的数组的两个连续元素之间的差在1到N的范围内。任务是使用给定的差异确定原始数组。如果可能，打印数组，否则打印-1 。
例子：
Input: diff[] = {2, -3, 2}
Output: arr[] = {2, 4, 1, 3}
4 – 2 = 2
1 – 4 = -3
3 – 1 = 2
Input: diff[] = {2, 2, 2}
Output: -1
方法：由于我们要生成范围为[1，n]的排列，并且每个数字只能出现一次。举个例子
arr[] = {2, -3, 2}
Here, P2 – P1 = 2, P3 – P2 = -3, P4 – P3 = 2.
Let P1 = x then P2 = x + 2, P3 = P2 – 3 = x + 2 – 3 = x – 1, P4 = P3 + 2 = x – 1 + 2 = x + 1.
So, P1 = x, P2 = x + 2, P3 = x – 1, P4 = x + 1.
Now since we want a permutation from 1 to n, therfore the P[i]’s we get above must also satisfy the condition. Since the value must be satisfied for each and every x, hence for our simplicity we take x = 0.
Now, P1 = 0, P2 = 2, P3 = -1, P4 = 1.
在对每个元素之间的连续差进行排序之后，我们将对p [i]进行排序，如果它们之间的差必须等于1。如果在任何索引处，我们都遇到一个元素p [i] ，使得p [i] – p [i – 1] ！= 1 ，则无法生成排列。为了生成最终的排列，我们将跟踪索引，我们可以使用map或unordered_map来进行。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to print the required permutation
void findPerm(int n, vector& differences)
{
vector ans;
ans.clear();
ans.push_back(0);
// Take x = 0 for simplicity
int x = 0;
// Calculate aint the differences
// and store it in a vector
for (int i = 0; i <= n - 2; ++i) {
int diff = differences[i];
x = x + diff;
ans.push_back(x);
}
// Preserve the original array
vector anss = ans;
sort(ans.begin(), ans.end());
int flag = -1;
// Check if aint the consecutive elements
// have difference = 1
for (int i = 1; i <= n - 1; ++i) {
int res = ans[i] - ans[i - 1];
if (res != 1) {
flag = 0;
}
}
// If consecutive elements don't have
// difference 1 at any position then
// the answer is impossible
if (flag == 0) {
cout << -1;
return;
}
// Else store the indices and values
// at those indices in a map
// and finainty print them
else {
unordered_map mpp;
mpp.clear();
int j = 1;
vector value_at_index;
for (auto& x : ans) {
mpp[x] = j;
++j;
}
for (auto& x : anss) {
value_at_index.push_back(mpp[x]);
}
for (auto& x : value_at_index) {
cout << x << " ";
}
cout << endl;
}
}
// Driver code
int main()
{
vector differences;
differences.push_back(2);
differences.push_back(-3);
differences.push_back(2);
int n = differences.size() + 1;
findPerm(n, differences);
return 0;
} Java
// Java program to implement the above approach
import java.util.*;
class GFG
{
// Function to print the required permutation
static void findPerm(int n, Vector differences)
{
Vector ans = new Vector();
ans.clear();
ans.add(0);
// Take x = 0 for simplicity
int x = 0;
// Calculate aint the differences
// and store it in a vector
for (int i = 0; i <= n - 2; ++i)
{
int diff = differences.get(i);
x = x + diff;
ans.add(x);
}
// Preserve the original array
Vector anss = new Vector();
for(Integer obj:ans)
anss.add(obj);
Collections.sort(ans);
int flag = -1;
// Check if aint the consecutive elements
// have difference = 1
for (int i = 1; i <= n - 1; ++i)
{
int res = ans.get(i) - ans.get(i - 1);
if (res != 1)
{
flag = 0;
}
}
// If consecutive elements don't have
// difference 1 at any position then
// the answer is impossible
if (flag == 0)
{
System.out.print(-1);
return;
}
// Else store the indices and values
// at those indices in a map
// and finainty print them
else
{
Map mpp = new HashMap<>();
mpp.clear();
int j = 1;
Vector value_at_index = new Vector();
for (Integer x1 : ans)
{
mpp.put(x1, j);
++j;
}
for (Integer x2 : anss)
{
value_at_index.add(mpp.get(x2));
}
for (Integer x3 : value_at_index)
{
System.out.print(x3 + " ");
}
System.out.println();
}
}
// Driver code
public static void main(String[] args)
{
Vector differences = new Vector();
differences.add(2);
differences.add(-3);
differences.add(2);
int n = differences.size() + 1;
findPerm(n, differences);
}
}
// This code is contributed by 29AjayKumar Python3
# Python3 implementation of the approach
# Function to print the required permutation
def findPerm(n, differences):
ans = []
ans.append(0)
# Take x = 0 for simplicity
x = 0
# Calculate athe differences
# and store it in a vector
for i in range(n - 1):
diff = differences[i]
x = x + diff
ans.append(x)
# Preserve the original array
anss = ans
ans = sorted(ans)
flag = -1
# Check if athe consecutive elements
# have difference = 1
for i in range(1, n):
res = ans[i] - ans[i - 1]
if (res != 1):
flag = 0
# If consecutive elements don't have
# difference 1 at any position then
# the answer is impossible
if (flag == 0):
print("-1")
return
# Else store the indices and values
# at those indices in a map
# and finainty print them
else:
mpp = dict()
j = 1
value_at_index = []
for x in ans:
mpp[x] = j
j += 1
for x in anss:
value_at_index.append(mpp[x])
for x in value_at_index:
print(x, end = " ")
print()
# Driver code
differences=[]
differences.append(2)
differences.append(-3)
differences.append(2)
n = len(differences) + 1
findPerm(n, differences)
# This code is contributed by mohit kumarC#
// C# program to implement the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to print the required permutation
static void findPerm(int n, List differences)
{
List ans = new List();
ans.Clear();
ans.Add(0);
// Take x = 0 for simplicity
int x = 0;
// Calculate aint the differences
// and store it in a vector
for(int i = 0; i <= n - 2; ++i)
{
int diff = differences[i];
x = x + diff;
ans.Add(x);
}
// Preserve the original array
List anss = new List();
foreach(int obj in ans)
anss.Add(obj);
ans.Sort();
int flag = -1;
// Check if aint the consecutive elements
// have difference = 1
for(int i = 1; i <= n - 1; ++i)
{
int res = ans[i] - ans[i - 1];
if (res != 1)
{
flag = 0;
}
}
// If consecutive elements don't have
// difference 1 at any position then
// the answer is impossible
if (flag == 0)
{
Console.Write(-1);
return;
}
// Else store the indices and values
// at those indices in a map
// and finainty print them
else
{
Dictionary mpp = new Dictionary();
mpp.Clear();
int j = 1;
List value_at_index = new List();
foreach (int x1 in ans)
{
mpp.Add(x1, j);
++j;
}
foreach (int x2 in anss)
{
value_at_index.Add(mpp[x2]);
}
foreach (int x3 in value_at_index)
{
Console.Write(x3 + " ");
}
Console.WriteLine();
}
}
// Driver code
public static void Main(String[] args)
{
List differences = new List();
differences.Add(2);
differences.Add(-3);
differences.Add(2);
int n = differences.Count + 1;
findPerm(n, differences);
}
}
// This code is contributed by Amit Katiyar 输出：
2 4 1 3