给定一个n个唯一整数的数组，其中数组中的每个元素都在[1，n]范围内。数组具有所有不同的元素，数组的大小为(n-2)。因此，此数组中缺少范围中的两个数字。找到两个缺失的数字。
例子：
Input : arr[] = {1, 3, 5, 6}, n = 6
Output : 2 4
Input : arr[] = {1, 2, 4}, n = 5
Output : 3 5
Input : arr[] = {1, 2}, n = 4
Output : 3 4
查找两个缺失的数字|第一组(一个有趣的线性时间解决方案)
在上面的文章中，我们讨论了解决此问题的两种方法。方法1需要O(n)多余的空间，而方法2可能导致溢出。在这篇文章中，讨论了一个新的解决方案。此处讨论的解决方案是O(n)时间，O(1)多余空间，不会引起溢出。
以下是步骤。
- 查找所有数组元素和从1到n的自然数的XOR。令数组为arr [] = {1、3、5、6}
XOR = (1 ^ 3 ^ 5 ^ 6) ^ (1 ^ 2 ^ 3 ^ 4 ^ 5 ^ 6) - 根据XOR的属性，相同的元素将被抵消，剩下的2 XOR 4 = 6(110)。但是我们不知道确切的数字，请让它们分别为X和Y。
- 仅当X和Y中的相应位不同时，才在xor中设置一个位。这是理解的关键步骤。
- 我们对XOR进行了设定。让我们考虑XOR中最右边的置位set_bit_no = 010
- 再一次，如果我们对arr []和1到n的所有具有最右边位的元素进行异或，我们将得到一个重复数，例如x。
Ex: Elements in arr[] with bit set: {3, 6} Elements from 1 to n with bit set {2, 3, 6} Result of XOR'ing all these is x = 2. - 类似地，如果我们对arr []和1到n中所有未设置最右位的元素进行异或，我们将得到另一个元素y。
Ex: Elements in arr[] with bit not set: {1, 5} Elements from 1 to n with bit not set {1, 4, 5} Result of XOR'ing all these is y = 4
下面是上述步骤的实现。
C++
// C++ Program to find 2 Missing Numbers using O(1)
// extra space and no overflow.
#include
// Function to find two missing numbers in range
// [1, n]. This function assumes that size of array
// is n-2 and all array elements are distinct
void findTwoMissingNumbers(int arr[], int n)
{
/* Get the XOR of all elements in arr[] and
{1, 2 .. n} */
int XOR = arr[0];
for (int i = 1; i < n-2; i++)
XOR ^= arr[i];
for (int i = 1; i <= n; i++)
XOR ^= i;
// Now XOR has XOR of two missing elements. Any set
// bit in it must be set in one missing and unset in
// other missing number
// Get a set bit of XOR (We get the rightmost set bit)
int set_bit_no = XOR & ~(XOR-1);
// Now divide elements in two sets by comparing rightmost
// set bit of XOR with bit at same position in each element.
int x = 0, y = 0; // Initialize missing numbers
for (int i = 0; i < n-2; i++)
{
if (arr[i] & set_bit_no)
x = x ^ arr[i]; /*XOR of first set in arr[] */
else
y = y ^ arr[i]; /*XOR of second set in arr[] */
}
for (int i = 1; i <= n; i++)
{
if (i & set_bit_no)
x = x ^ i; /* XOR of first set in arr[] and
{1, 2, ...n }*/
else
y = y ^ i; /* XOR of second set in arr[] and
{1, 2, ...n } */
}
printf("Two Missing Numbers are\n %d %d", x, y);
}
// Driver program to test above function
int main()
{
int arr[] = {1, 3, 5, 6};
// Range of numbers is 2 plus size of array
int n = 2 + sizeof(arr)/sizeof(arr[0]);
findTwoMissingNumbers(arr, n);
return 0;
} Java
// Java Program to find 2 Missing Numbers
import java.util.*;
class GFG {
// Function to find two missing numbers in range
// [1, n]. This function assumes that size of array
// is n-2 and all array elements are distinct
static void findTwoMissingNumbers(int arr[], int n)
{
/* Get the XOR of all elements in arr[] and
{1, 2 .. n} */
int XOR = arr[0];
for (int i = 1; i < n-2; i++)
XOR ^= arr[i];
for (int i = 1; i <= n; i++)
XOR ^= i;
// Now XOR has XOR of two missing elements.
// Any set bit in it must be set in one missing
// and unset in other missing number
// Get a set bit of XOR (We get the rightmost
// set bit)
int set_bit_no = XOR & ~(XOR-1);
// Now divide elements in two sets by comparing
// rightmost set bit of XOR with bit at same
// position in each element.
int x = 0, y = 0; // Initialize missing numbers
for (int i = 0; i < n-2; i++)
{
if ((arr[i] & set_bit_no) > 0)
/*XOR of first set in arr[] */
x = x ^ arr[i];
else
/*XOR of second set in arr[] */
y = y ^ arr[i];
}
for (int i = 1; i <= n; i++)
{
if ((i & set_bit_no)>0)
/* XOR of first set in arr[] and
{1, 2, ...n }*/
x = x ^ i;
else
/* XOR of second set in arr[] and
{1, 2, ...n } */
y = y ^ i;
}
System.out.println("Two Missing Numbers are ");
System.out.println( x + " " + y);
}
/* Driver program to test above function */
public static void main(String[] args)
{
int arr[] = {1, 3, 5, 6};
// Range of numbers is 2 plus size of array
int n = 2 +arr.length;
findTwoMissingNumbers(arr, n);
}
}
// This code is contributed by Arnav Kr. Mandal.Python3
# Python Program to find 2 Missing
# Numbers using O(1)
# extra space and no overflow.
# Function to find two missing
# numbers in range
# [1, n]. This function assumes
# that size of array
# is n-2 and all array elements
# are distinct
def findTwoMissingNumbers(arr, n):
# Get the XOR of all
# elements in arr[] and
# {1, 2 .. n}
XOR = arr[0]
for i in range(1,n-2):
XOR ^= arr[i]
for i in range(1,n+1):
XOR ^= i
# Now XOR has XOR of two
# missing elements. Any set
# bit in it must be set in
# one missing and unset in
# other missing number
# Get a set bit of XOR
# (We get the rightmost set bit)
set_bit_no = XOR & ~(XOR-1)
# Now divide elements in two sets
# by comparing rightmost
# set bit of XOR with bit at same
# position in each element.
x = 0
# Initialize missing numbers
y = 0
for i in range(0,n-2):
if arr[i] & set_bit_no:
# XOR of first set in arr[]
x = x ^ arr[i]
else:
# XOR of second set in arr[]
y = y ^ arr[i]
for i in range(1,n+1):
if i & set_bit_no:
# XOR of first set in arr[] and
# {1, 2, ...n }
x = x ^ i
else:
# XOR of second set in arr[] and
# {1, 2, ...n }
y = y ^ i
print ("Two Missing Numbers are\n%d %d"%(x,y))
# Driver program to test
# above function
arr = [1, 3, 5, 6]
# Range of numbers is 2
# plus size of array
n = 2 + len(arr)
findTwoMissingNumbers(arr, n)
# This code is contributed
# by Shreyanshi Arun.C#
// Program to find 2 Missing Numbers
using System;
class GFG {
// Function to find two missing
// numbers in range [1, n].This
// function assumes that size of
// array is n-2 and all array
// elements are distinct
static void findTwoMissingNumbers(int[] arr, int n)
{
// Get the XOR of all elements
// in arr[] and {1, 2 .. n}
int XOR = arr[0];
for (int i = 1; i < n - 2; i++)
XOR ^= arr[i];
for (int i = 1; i <= n; i++)
XOR ^= i;
// Now XOR has XOR of two missing
// element. Any set bit in it must
// be set in one missing and unset
// in other missing number
// Get a set bit of XOR (We get the
// rightmost set bit)
int set_bit_no = XOR & ~(XOR - 1);
// Now divide elements in two sets
// by comparing rightmost set bit
// of XOR with bit at same position
// in each element.
int x = 0, y = 0;
// Initialize missing numbers
for (int i = 0; i < n - 2; i++) {
if ((arr[i] & set_bit_no) > 0)
// XOR of first set in arr[]
x = x ^ arr[i];
else
// XOR of second set in arr[]
y = y ^ arr[i];
}
for (int i = 1; i <= n; i++) {
if ((i & set_bit_no) > 0)
// XOR of first set in arr[]
// and {1, 2, ...n }
x = x ^ i;
else
// XOR of second set in arr[]
// and {1, 2, ...n }
y = y ^ i;
}
Console.WriteLine("Two Missing Numbers are ");
Console.WriteLine(x + " " + y);
}
// Driver program
public static void Main()
{
int[] arr = { 1, 3, 5, 6 };
// Range of numbers is 2 plus
// size of array
int n = 2 + arr.Length;
findTwoMissingNumbers(arr, n);
}
}
// This code is contributed by Anant Agarwal.PHP
输出：
Two Missing Numbers are
2 4
时间复杂度：O(n)
辅助空间：O(1)
没有整数溢出