给定一个由N个整数组成的数组A []并且数组B []的第一个元素为K ,任务是从A []构造数组B [] ,使得对于任何索引i , A [i]都是按位的B []除B [i]之外的所有数组元素的XOR。
例子:
Input: A[] = {13, 14, 10, 6}, K = 2
Output: 2 1 5 9
Explanation:
For any index i, A[i] is the Bitwise XOR of all elements of B[] except B[i].
- B[1] ^ B[2] ^ B[3] = 1 ^ 5 ^ 9 = 13 = A[0]
- B[0] ^ B[2] ^ B[3] = 2 ^ 5 ^ 9 = 14 = A[1]
- B[0] ^ B[1] ^ B[3] = 2 ^ 1 ^ 9 = 10 = A[2]
- B[0] ^ B[1] ^ B[2] = 2 ^ 1 ^ 5 = 6 = A[3]
Input: A[] = {3, 5, 0, 2, 4}, K = 2
Output: 2 4 1 3 5
方法:这个想法是基于观察到的相同值偶数计算的按位XOR为0 。
For any index i,
A[i] = B[0] ^ B[1] ^ … B[i-1] ^ B[i+1] ^ … B[n-1]
Therefore, XOR of all elements of B[], totalXor = B[0] ^ B[1] ^ … B[i – 1] ^ B[i] ^ B[i + 1] ^ … ^ B[N – 1].
Therefore, B[i] = totalXor ^ A[i]. (Since every element occurs twice except B[i])
请按照以下步骤解决问题:
- 将数组B []中存在的所有元素的按位XOR存储在一个变量中,例如totalXOR ,其中totalXOR = A [0] ^ K。
- 遍历每个数组元素A [i]的给定数组A [] ,将B [i]的值存储为totalXOR ^ A [i] 。
- 完成上述步骤后,打印存储在数组B []中的元素。
下面是上述方法的实现:
C++
// C++ program for the above aproach
#include
using namespace std;
// Function to construct an array
// with each element equal to XOR
// of all array elements except
// the element at the same index
void constructArray(int A[], int N,
int K)
{
// Original array
int B[N];
// Stores Bitwise XOR of array
int totalXOR = A[0] ^ K;
// Calculate XOR of all array elements
for (int i = 0; i < N; i++)
B[i] = totalXOR ^ A[i];
// Print the original array B[]
for (int i = 0; i < N; i++) {
cout << B[i] << " ";
}
}
// Driver Code
int main()
{
int A[] = { 13, 14, 10, 6 }, K = 2;
int N = sizeof(A) / sizeof(A[0]);
// Function Call
constructArray(A, N, K);
return 0;
}
Java
// Java program for the above aproach
class GFG{
// Function to construct an array
// with each element equal to XOR
// of all array elements except
// the element at the same index
static void constructArray(int A[], int N,
int K)
{
// Original array
int B[] = new int[N];
// Stores Bitwise XOR of array
int totalXOR = A[0] ^ K;
// Calculate XOR of all array elements
for(int i = 0; i < N; i++)
B[i] = totalXOR ^ A[i];
// Print the original array B[]
for(int i = 0; i < N; i++)
{
System.out.print(B[i] + " ");
}
}
// Driver Code
public static void main(String[] args)
{
int A[] = { 13, 14, 10, 6 }, K = 2;
int N = A.length;
// Function Call
constructArray(A, N, K);
}
}
// This code is contributed by divyeshrabadiya07
Python3
# Python program for the above aproach
# Function to construct an array
# with each element equal to XOR
# of all array elements except
# the element at the same index
def constructArray(A, N, K):
# Original array
B = [0] * N;
# Stores Bitwise XOR of array
totalXOR = A[0] ^ K;
# Calculate XOR of all array elements
for i in range(N):
B[i] = totalXOR ^ A[i];
# Prthe original array B
for i in range(N):
print(B[i], end = " ");
# Driver Code
if __name__ == '__main__':
A = [13, 14, 10, 6];
K = 2;
N = len(A);
# Function Call
constructArray(A, N, K);
# This code is contributed by Princi Singh
C#
// C# program for the above aproach
using System;
using System.Collections;
class GFG {
// Function to construct an array
// with each element equal to XOR
// of all array elements except
// the element at the same index
static void constructArray(int[] A, int N,
int K)
{
// Original array
int[] B = new int[N];
// Stores Bitwise XOR of array
int totalXOR = A[0] ^ K;
// Calculate XOR of all array elements
for(int i = 0; i < N; i++)
B[i] = totalXOR ^ A[i];
// Print the original array B[]
for(int i = 0; i < N; i++)
{
Console.Write(B[i] + " ");
}
}
static void Main() {
int[] A = { 13, 14, 10, 6 };
int K = 2;
int N = A.Length;
// Function Call
constructArray(A, N, K);
}
}
// This code is contributed by divyesh072019
Javascript
输出:
2 1 5 9
时间复杂度: O(N)
辅助空间: O(1)