给定一个数组,其中所有元素除两次外均出现偶数次,请打印出两个奇数出现的元素。可以假设数组的大小至少为2。
例子:
Input : arr[] = {2, 3, 8, 4, 4, 3, 7, 8}
Output : 2 7
Input : arr[] = {15, 10, 10, 50 7, 5, 5, 50, 50, 50, 50, 50}
Output : 7 15
一个简单的解决方案是使用两个嵌套循环。外循环遍历所有元素。内部循环对当前元素的出现进行计数。我们打印出现次数为奇数的元素。如果此解决方案为O(n 2 ),则时间复杂度
更好的解决方案是使用哈希。该解决方案的时间复杂度为O(n),但需要额外的空间。
一个有效的解决方案是使用按位运算运算符。这个想法是基于在两个缺失要素和两个重复要素中使用的方法。
C++
// CPP code to find two odd occurring elements
// in an array where all other elements appear
// even number of times.
#include
using namespace std;
void printOdds(int arr[], int n)
{
// Find XOR of all numbers
int res = 0;
for (int i = 0; i < n; i++)
res = res ^ arr[i];
// Find a set bit in the XOR (We find
// rightmost set bit here)
int set_bit = res & (~(res - 1));
// Traverse through all numbers and
// divide them in two groups
// (i) Having set bit set at same
// position as the only set bit
// in set_bit
// (ii) Having 0 bit at same position
// as the only set bit in set_bit
int x = 0, y = 0;
for (int i = 0; i < n; i++) {
if (arr[i] & set_bit)
x = x ^ arr[i];
else
y = y ^ arr[i];
}
// XOR of two different sets are our
// required numbers.
cout << x << " " << y;
}
// Driver code
int main()
{
int arr[] = { 2, 3, 3, 4, 4, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
printOdds(arr, n);
return 0;
}
Java
// Java code to find two
// odd occurring elements
// in an array where all
// other elements appear
// even number of times.
class GFG
{
static void printOdds(int arr[],
int n)
{
// Find XOR of
// all numbers
int res = 0;
for (int i = 0; i < n; i++)
res = res ^ arr[i];
// Find a set bit in the
// XOR (We find rightmost
// set bit here)
int set_bit = res &
(~(res - 1));
// Traverse through all
// numbers and divide them
// in two groups (i) Having
// set bit set at same position
// as the only set bit in
// set_bit (ii) Having 0 bit at
// same position as the only
// set bit in set_bit
int x = 0, y = 0;
for (int i = 0; i < n; i++)
{
if ((arr[i] & set_bit) != 0)
x = x ^ arr[i];
else
y = y ^ arr[i];
}
// XOR of two different
// sets are our required
// numbers.
System.out.println( x + " " + y);
}
// Driver code
public static void main(String [] args)
{
int arr[] = { 2, 3, 3,
4, 4, 5 };
int n = arr.length;
printOdds(arr, n);
}
}
// This code is contributed by
// Smitha Dinesh Semwal
Python3
# Python 3 code to find two
# odd occurring elements in
# an array where all other
# elements appear even number
# of times.
def printOdds(arr, n):
# Find XOR of all numbers
res = 0
for i in range(0, n):
res = res ^ arr[i]
# Find a set bit in
# the XOR (We find
# rightmost set bit here)
set_bit = res & (~(res - 1))
# Traverse through all numbers
# and divide them in two groups
# (i) Having set bit set at
# same position as the only set
# bit in set_bit
# (ii) Having 0 bit at same
# position as the only set
# bit in set_bit
x = 0
y = 0
for i in range(0, n):
if (arr[i] & set_bit):
x = x ^ arr[i]
else:
y = y ^ arr[i]
# XOR of two different
# sets are our
# required numbers.
print(x , y, end = "")
# Driver code
arr = [2, 3, 3, 4, 4, 5 ]
n = len(arr)
printOdds(arr, n)
# This code is contributed
# by Smitha
C#
// C# code to find two
// odd occurring elements
// in an array where all
// other elements appear
// even number of times.
using System;
class GFG
{
static void printOdds(int []arr,
int n)
{
// Find XOR of
// all numbers
int res = 0;
for (int i = 0; i < n; i++)
res = res ^ arr[i];
// Find a set bit in the
// XOR (We find rightmost
// set bit here)
int set_bit = res &
(~(res - 1));
// Traverse through all
// numbers and divide them
// in two groups (i) Having
// set bit set at same position
// as the only set bit in
// set_bit (ii) Having 0 bit at
// same position as the only
// set bit in set_bit
int x = 0, y = 0;
for (int i = 0; i < n; i++)
{
if ((arr[i] & set_bit) != 0)
x = x ^ arr[i];
else
y = y ^ arr[i];
}
// XOR of two different
// sets are our required
// numbers.
Console.WriteLine(x + " " + y);
}
// Driver code
public static void Main()
{
int []arr = { 2, 3, 3,
4, 4, 5 };
int n = arr.Length;
printOdds(arr, n);
}
}
// This code is contributed by
// Akanksha Rai(Abby_akku)
PHP
Javascript
输出:
5 2
时间复杂度: O(n)
辅助空间: O(1)