给定一个由N-1 个元素组成的序列arr[] ，它是数组中所有相邻对的异或，任务是从arr[] 中找到原始数组。
注意：假设 N 总是奇数并且 arr[] 包含 N 个自然数的排列。
例子：
Input: arr[] = {3, 1}
Output: 1 2 3
Explanation:
The XOR of the output array will lead to the given array that is:
1 ^ 2 = 3
2 ^ 3 = 1
Input: arr[] = {7, 5, 3, 7}
Output: 3 4 1 2 5
Explanation:
The XOR of the output array will lead to the given array that is:
3 ^ 4 = 7
4 ^ 1 = 5
1 ^ 2 = 3
2 ^ 5 = 7
方法：
- 思路是求1到N的所有元素的异或和给定数组相邻元素的异或，求出期望数组的最后一个元素。
- 由于相邻元素的 XOR 将包含除最后一个元素之外的所有元素，因此 this 与从 1 到 N 的所有数字的 XOR 将给出预期排列的最后一个元素。
例如：
Let's the expected array be - {a, b, c, d, e}
Then the XOR array for this array will be -
{a^b, b^c, c^d, d^e}
Now XOR of all the element from 1 to N -
xor_all => a ^ b ^ c ^ d ^ e
XOR of the adjacent elements -
xor_adjacent => ((a ^ b) ^ (c ^ d))
Now the XOR of the both the array will be the
last element of the expected permutation
=> (a ^ b ^ c ^ d ^ e) ^ ((a ^ b) ^ (c ^ d))
=> As all elements are in pair except the last element.
=> (a ^ a ^ b ^ b ^ c ^ c ^ d ^ d ^ e)
=> (0 ^ 0 ^ 0 ^ 0 ^ e)
=> e- 现在对于元素的其余部分，连续地，在最后一个元素的异或上，我们将获得最后一个第二个元素，即d 。
- 反复更新最后一个元素，最终得到第一个元素，即a 。
下面是上述方法的实现：
C++
// C++ implementation to find the
// Array from the XOR array
// of the adjacent elements of array
#include
using namespace std;
// XOR of all elements from 1 to N
int xor_all_elements(int n)
{
switch (n & 3) {
case 0:
return n;
case 1:
return 1;
case 2:
return n + 1;
case 3:
return 0;
}
}
// Function to find the Array
// from the XOR Array
vector findArray(int xorr[], int n)
{
// Take a vector to store
// the permutation
vector arr;
// XOR of N natural numbers
int xor_all = xor_all_elements(n);
int xor_adjacent = 0;
// Loop to find the XOR of
// adjacent elements of the XOR Array
for (int i = 0; i < n - 1; i += 2) {
xor_adjacent = xor_adjacent ^ xorr[i];
}
int last_element = xor_all ^ xor_adjacent;
arr.push_back(last_element);
// Loop to find the other
// elements of the permutation
for (int i = n - 2; i >= 0; i--) {
// Finding the next and next elements
last_element = xorr[i] ^ last_element;
arr.push_back(last_element);
}
return arr;
}
// Driver Code
int main()
{
vector arr;
int xorr[] = { 7, 5, 3, 7 };
int n = 5;
arr = findArray(xorr, n);
// Required Permutation
for (int i = n - 1; i >= 0; i--) {
cout << arr[i] << " ";
}
} Java
// Java implementation to find the
// Array from the XOR array
// of the adjacent elements of array
import java.util.*;
class GFG{
// XOR of all elements from 1 to N
static int xor_all_elements(int n)
{
switch (n & 3) {
case 0:
return n;
case 1:
return 1;
case 2:
return n + 1;
}
return 0;
}
// Function to find the Array
// from the XOR Array
static Vector findArray(int xorr[], int n)
{
// Take a vector to store
// the permutation
Vector arr = new Vector();
// XOR of N natural numbers
int xor_all = xor_all_elements(n);
int xor_adjacent = 0;
// Loop to find the XOR of
// adjacent elements of the XOR Array
for (int i = 0; i < n - 1; i += 2) {
xor_adjacent = xor_adjacent ^ xorr[i];
}
int last_element = xor_all ^ xor_adjacent;
arr.add(last_element);
// Loop to find the other
// elements of the permutation
for (int i = n - 2; i >= 0; i--)
{
// Finding the next and next elements
last_element = xorr[i] ^ last_element;
arr.add(last_element);
}
return arr;
}
// Driver Code
public static void main(String[] args)
{
Vector arr = new Vector();
int xorr[] = { 7, 5, 3, 7 };
int n = 5;
arr = findArray(xorr, n);
// Required Permutation
for (int i = n - 1; i >= 0; i--)
{
System.out.print(arr.get(i)+ " ");
}
}
}
// This code is contributed by PrinciRaj1992 Python3
# Python3 implementation to find the
# Array from the XOR array
# of the adjacent elements of array
# XOR of all elements from 1 to N
def xor_all_elements(n):
if n & 3 == 0:
return n
elif n & 3 == 1:
return 1
elif n & 3 == 2:
return n + 1
else:
return 0
# Function to find the Array
# from the XOR Array
def findArray(xorr, n):
# Take a vector to store
# the permutation
arr = []
# XOR of N natural numbers
xor_all = xor_all_elements(n)
xor_adjacent = 0
# Loop to find the XOR of
# adjacent elements of the XOR Array
for i in range(0, n - 1, 2):
xor_adjacent = xor_adjacent ^ xorr[i]
last_element = xor_all ^ xor_adjacent
arr.append(last_element)
# Loop to find the other
# elements of the permutation
for i in range(n - 2, -1, -1):
# Finding the next and next elements
last_element = xorr[i] ^ last_element
arr.append(last_element)
return arr
# Driver Code
xorr = [7, 5, 3, 7]
n = 5
arr = findArray(xorr, n)
# Required Permutation
for i in range(n - 1, -1, -1):
print(arr[i], end=" ")
# This code is contributed by mohit kumar 29C#
// C# implementation to find the
// Array from the XOR array
// of the adjacent elements of array
using System;
using System.Collections.Generic;
class GFG{
// XOR of all elements from 1 to N
static int xor_all_elements(int n)
{
switch (n & 3) {
case 0:
return n;
case 1:
return 1;
case 2:
return n + 1;
}
return 0;
}
// Function to find the Array
// from the XOR Array
static List findArray(int []xorr, int n)
{
// Take a vector to store
// the permutation
List arr = new List();
// XOR of N natural numbers
int xor_all = xor_all_elements(n);
int xor_adjacent = 0;
// Loop to find the XOR of
// adjacent elements of the XOR Array
for (int i = 0; i < n - 1; i += 2) {
xor_adjacent = xor_adjacent ^ xorr[i];
}
int last_element = xor_all ^ xor_adjacent;
arr.Add(last_element);
// Loop to find the other
// elements of the permutation
for (int i = n - 2; i >= 0; i--)
{
// Finding the next and next elements
last_element = xorr[i] ^ last_element;
arr.Add(last_element);
}
return arr;
}
// Driver Code
public static void Main(String[] args)
{
List arr = new List();
int []xorr = { 7, 5, 3, 7 };
int n = 5;
arr = findArray(xorr, n);
// Required Permutation
for (int i = n - 1; i >= 0; i--)
{
Console.Write(arr[i]+ " ");
}
}
}
// This code contributed by Rajput-Ji Javascript
输出：
3 4 1 2 5性能分析：
- 时间复杂度：在上面的方法中，我们遍历整个异或数组以找到相邻元素的异或，那么最坏情况下的复杂度将是O(N)
- 空间复杂度：在上面的方法中，有一个向量数组用于存储从 1 到 N 的数字的排列，那么空间复杂度将为O(N)