给定数组中缺少偶数和奇数元素
给定两个整数数组，偶[] 和奇[] ，它们分别包含连续的偶和奇元素，每个数组中缺少一个元素。任务是找到丢失的元素。
示例:
输入:偶数[] = {6，4，8，14，10}，奇数[] = {7，5，3，11，13} 输出: 偶数= 12 奇数= 9
输入:偶[] = {4，6，2，10}，奇[] = {7，5，9，13，11，17} 输出: 偶= 8 奇= 15
方法:将偶[] 数组中的最小偶元素和最大偶元素存储在变量 minEven 和 maxEven 中。我们知道，前 N 个偶数的和为 N * (N + 1) 。计算从 2 到的偶数之和我甚至说 sum1 和从 2 到 maxEven 说 sum2 的偶数之和。现在，偶数数组所需的和将是req sum = sum 2–sum 1+minEven，从这个 reqSum 中减去even【】数组的和将会给出缺失的偶数。 同样，缺失的奇数也可以找到，因为我们知道前 N 个奇数的和为 N 2 。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the missing numbers
void findMissingNums(int even[], int sizeEven, int odd[], int sizeOdd)
{
// To store the minimum and the maximum
// odd and even elements from the arrays
int minEven = INT_MAX;
int maxEven = INT_MIN;
int minOdd = INT_MAX;
int maxOdd = INT_MIN;
// To store the sum of the array elements
int sumEvenArr = 0, sumOddArr = 0;
// Get the minimum and the maximum
// even elements from the array
for (int i = 0; i < sizeEven; i++) {
minEven = min(minEven, even[i]);
maxEven = max(maxEven, even[i]);
sumEvenArr += even[i];
}
// Get the minimum and the maximum
// odd elements from the array
for (int i = 0; i < sizeOdd; i++) {
minOdd = min(minOdd, odd[i]);
maxOdd = max(maxOdd, odd[i]);
sumOddArr += odd[i];
}
// To store the total terms in the series
// and the required sum of the array
int totalTerms = 0, reqSum = 0;
// Total terms from 2 to minEven
totalTerms = minEven / 2;
// Sum of all even numbers from 2 to minEven
int evenSumMin = totalTerms * (totalTerms + 1);
// Total terms from 2 to maxEven
totalTerms = maxEven / 2;
// Sum of all even numbers from 2 to maxEven
int evenSumMax = totalTerms * (totalTerms + 1);
// Required sum for the even array
reqSum = evenSumMax - evenSumMin + minEven;
// Missing even number
cout << "Even = " << reqSum - sumEvenArr << "\n";
// Total terms from 1 to minOdd
totalTerms = (minOdd / 2) + 1;
// Sum of all odd numbers from 1 to minOdd
int oddSumMin = totalTerms * totalTerms;
// Total terms from 1 to maxOdd
totalTerms = (maxOdd / 2) + 1;
// Sum of all odd numbers from 1 to maxOdd
int oddSumMax = totalTerms * totalTerms;
// Required sum for the odd array
reqSum = oddSumMax - oddSumMin + minOdd;
// Missing odd number
cout << "Odd = " << reqSum - sumOddArr;
}
// Driver code
int main()
{
int even[] = { 6, 4, 8, 14, 10 };
int sizeEven = sizeof(even) / sizeof(even[0]);
int odd[] = { 7, 5, 3, 11, 13 };
int sizeOdd = sizeof(odd) / sizeof(odd[0]);
findMissingNums(even, sizeEven, odd, sizeOdd);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG
{
// Function to find the missing numbers
static void findMissingNums(int even[], int sizeEven,
int odd[], int sizeOdd)
{
// To store the minimum and the maximum
// odd and even elements from the arrays
int minEven =Integer.MAX_VALUE;
int maxEven =Integer.MIN_VALUE;
int minOdd = Integer.MAX_VALUE;
int maxOdd = Integer.MIN_VALUE;
// To store the sum of the array elements
int sumEvenArr = 0, sumOddArr = 0;
// Get the minimum and the maximum
// even elements from the array
for (int i = 0; i < sizeEven; i++)
{
minEven = Math.min(minEven, even[i]);
maxEven = Math.max(maxEven, even[i]);
sumEvenArr += even[i];
}
// Get the minimum and the maximum
// odd elements from the array
for (int i = 0; i < sizeOdd; i++)
{
minOdd = Math.min(minOdd, odd[i]);
maxOdd = Math.max(maxOdd, odd[i]);
sumOddArr += odd[i];
}
// To store the total terms in the series
// and the required sum of the array
int totalTerms = 0, reqSum = 0;
// Total terms from 2 to minEven
totalTerms = minEven / 2;
// Sum of all even numbers
// from 2 to minEven
int evenSumMin = (totalTerms *
(totalTerms + 1));
// Total terms from 2 to maxEven
totalTerms = maxEven / 2;
// Sum of all even numbers from
// 2 to maxEven
int evenSumMax = (totalTerms *
(totalTerms + 1));
// Required sum for the even array
reqSum = evenSumMax - evenSumMin + minEven;
// Missing even number
System.out.println("Even = " +
(reqSum - sumEvenArr));
// Total terms from 1 to minOdd
totalTerms = (minOdd / 2) + 1;
// Sum of all odd numbers
// from 1 to minOdd
int oddSumMin = totalTerms * totalTerms;
// Total terms from 1 to maxOdd
totalTerms = (maxOdd / 2) + 1;
// Sum of all odd numbers
// from 1 to maxOdd
int oddSumMax = totalTerms * totalTerms;
// Required sum for the odd array
reqSum = oddSumMax - oddSumMin + minOdd;
// Missing odd number
System.out.println("Odd = " +
(reqSum - sumOddArr));
}
// Driver code
public static void main(String[] args)
{
int even[] = { 6, 4, 8, 14, 10 };
int sizeEven = even.length;
int odd[] = { 7, 5, 3, 11, 13 };
int sizeOdd = odd.length;
findMissingNums(even, sizeEven,
odd, sizeOdd);
}
}
// This code is contributed
// by Code_Mech.
Python 3
# Python3 implementation of the approach
import sys
# Function to find the missing numbers
def findMissingNums(even, sizeEven,
odd, sizeOdd):
# To store the minimum and the
# maximum odd and even elements
# from the arrays
minEven = sys.maxsize ;
maxEven = -(sys.maxsize - 1);
minOdd = sys.maxsize ;
maxOdd = -(sys.maxsize - 1);
# To store the sum of the
# array elements
sumEvenArr = 0;
sumOddArr = 0;
# Get the minimum and the maximum
# even elements from the array
for i in range(sizeEven) :
minEven = min(minEven, even[i]);
maxEven = max(maxEven, even[i]);
sumEvenArr += even[i];
# Get the minimum and the maximum
# odd elements from the array
for i in range(sizeOdd) :
minOdd = min(minOdd, odd[i]);
maxOdd = max(maxOdd, odd[i]);
sumOddArr += odd[i];
# To store the total terms in
# the series and the required
# sum of the array
totalTerms = 0;
reqSum = 0;
# Total terms from 2 to minEven
totalTerms = minEven // 2;
# Sum of all even numbers from
# 2 to minEven
evenSumMin = (totalTerms *
(totalTerms + 1));
# Total terms from 2 to maxEven
totalTerms = maxEven // 2;
# Sum of all even numbers from
# 2 to maxEven
evenSumMax = (totalTerms *
(totalTerms + 1));
# Required sum for the even array
reqSum = (evenSumMax -
evenSumMin + minEven);
# Missing even number
print("Even =", reqSum -
sumEvenArr);
# Total terms from 1 to minOdd
totalTerms = (minOdd // 2) + 1;
# Sum of all odd numbers from
# 1 to minOdd
oddSumMin = totalTerms * totalTerms;
# Total terms from 1 to maxOdd
totalTerms = (maxOdd // 2) + 1;
# Sum of all odd numbers from
# 1 to maxOdd
oddSumMax = totalTerms * totalTerms;
# Required sum for the odd array
reqSum = (oddSumMax -
oddSumMin + minOdd);
# Missing odd number
print("Odd =", reqSum - sumOddArr);
# Driver code
if __name__ == "__main__" :
even = [ 6, 4, 8, 14, 10 ];
sizeEven = len(even)
odd = [ 7, 5, 3, 11, 13 ];
sizeOdd = len(odd) ;
findMissingNums(even, sizeEven,
odd, sizeOdd);
# This code is contributed by Ryuga
C
// C# implementation of the approach
using System;
class GFG
{
// Function to find the missing numbers
static void findMissingNums(int []even, int sizeEven,
int []odd, int sizeOdd)
{
// To store the minimum and the maximum
// odd and even elements from the arrays
int minEven =int.MaxValue;
int maxEven =int.MinValue;
int minOdd = int.MaxValue;
int maxOdd = int.MinValue;
// To store the sum of the array elements
int sumEvenArr = 0, sumOddArr = 0;
// Get the minimum and the maximum
// even elements from the array
for (int i = 0; i < sizeEven; i++)
{
minEven = Math.Min(minEven, even[i]);
maxEven = Math.Max(maxEven, even[i]);
sumEvenArr += even[i];
}
// Get the minimum and the maximum
// odd elements from the array
for (int i = 0; i < sizeOdd; i++)
{
minOdd = Math.Min(minOdd, odd[i]);
maxOdd = Math.Max(maxOdd, odd[i]);
sumOddArr += odd[i];
}
// To store the total terms in the series
// and the required sum of the array
int totalTerms = 0, reqSum = 0;
// Total terms from 2 to minEven
totalTerms = minEven / 2;
// Sum of all even numbers
// from 2 to minEven
int evenSumMin = (totalTerms *
(totalTerms + 1));
// Total terms from 2 to maxEven
totalTerms = maxEven / 2;
// Sum of all even numbers from
// 2 to maxEven
int evenSumMax = (totalTerms *
(totalTerms + 1));
// Required sum for the even array
reqSum = evenSumMax - evenSumMin + minEven;
// Missing even number
Console.WriteLine("Even = " +
(reqSum - sumEvenArr));
// Total terms from 1 to minOdd
totalTerms = (minOdd / 2) + 1;
// Sum of all odd numbers
// from 1 to minOdd
int oddSumMin = totalTerms * totalTerms;
// Total terms from 1 to maxOdd
totalTerms = (maxOdd / 2) + 1;
// Sum of all odd numbers
// from 1 to maxOdd
int oddSumMax = totalTerms * totalTerms;
// Required sum for the odd array
reqSum = oddSumMax - oddSumMin + minOdd;
// Missing odd number
Console.WriteLine("Odd = " +
(reqSum - sumOddArr));
}
// Driver code
static void Main()
{
int []even = { 6, 4, 8, 14, 10 };
int sizeEven = even.Length;
int []odd = { 7, 5, 3, 11, 13 };
int sizeOdd = odd.Length;
findMissingNums(even, sizeEven,
odd, sizeOdd);
}
}
// This code is contributed
// by chandan_jnu
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function to find the missing numbers
function findMissingNums($even, $sizeEven,
$odd, $sizeOdd)
{
// To store the minimum and the maximum
// odd and even elements from the arrays
$minEven = PHP_INT_MAX;
$maxEven = PHP_INT_MIN;
$minOdd = PHP_INT_MAX;
$maxOdd = PHP_INT_MIN;
// To store the sum of the array elements
$sumEvenArr = $sumOddArr = 0;
// Get the minimum and the maximum
// even elements from the array
for ($i = 0; $i < $sizeEven; $i++)
{
$minEven = min($minEven, $even[$i]);
$maxEven = max($maxEven, $even[$i]);
$sumEvenArr += $even[$i];
}
// Get the minimum and the maximum
// odd elements from the array
for ($i = 0; $i < $sizeOdd; $i++)
{
$minOdd = min($minOdd, $odd[$i]);
$maxOdd = max($maxOdd, $odd[$i]);
$sumOddArr += $odd[$i];
}
// To store the total terms in the series
// and the required sum of the array
$totalTerms = $reqSum = 0;
// Total terms from 2 to minEven
$totalTerms = (int)($minEven / 2);
// Sum of all even numbers
// from 2 to minEven
$evenSumMin = $totalTerms *
($totalTerms + 1);
// Total terms from 2 to maxEven
$totalTerms = (int)($maxEven / 2);
// Sum of all even numbers
// from 2 to maxEven
$evenSumMax = $totalTerms *
($totalTerms + 1);
// Required sum for the even array
$reqSum = ($evenSumMax -
$evenSumMin + $minEven);
// Missing even number
echo "Even = " . ($reqSum -
$sumEvenArr) . "\n";
// Total terms from 1 to minOdd
$totalTerms = (int)(($minOdd / 2) + 1);
// Sum of all odd numbers
// from 1 to minOdd
$oddSumMin = $totalTerms * $totalTerms;
// Total terms from 1 to maxOdd
$totalTerms = (int)(($maxOdd / 2) + 1);
// Sum of all odd numbers
// from 1 to maxOdd
$oddSumMax = $totalTerms * $totalTerms;
// Required sum for the odd array
$reqSum = ($oddSumMax -
$oddSumMin + $minOdd);
// Missing odd number
echo "Odd = " . ($reqSum - $sumOddArr);
}
// Driver code
$even = array( 6, 4, 8, 14, 10 );
$sizeEven = count($even);
$odd = array( 7, 5, 3, 11, 13 );
$sizeOdd = count($odd);
findMissingNums($even, $sizeEven,
$odd, $sizeOdd);
// This code is contributed
// by chandan_jnu
?>
java 描述语言
<script>
// Javascript implementation of the approach
// Function to find the missing numbers
function findMissingNums(even, sizeEven, odd, sizeOdd)
{
// To store the minimum and the maximum
// odd and even elements from the arrays
let minEven = Number.MAX_VALUE;
let maxEven = Number.MIN_VALUE;
let minOdd = Number.MAX_VALUE;
let maxOdd = Number.MIN_VALUE;
// To store the sum of the array elements
let sumEvenArr = 0, sumOddArr = 0;
// Get the minimum and the maximum
// even elements from the array
for (let i = 0; i < sizeEven; i++)
{
minEven = Math.min(minEven, even[i]);
maxEven = Math.max(maxEven, even[i]);
sumEvenArr += even[i];
}
// Get the minimum and the maximum
// odd elements from the array
for (let i = 0; i < sizeOdd; i++)
{
minOdd = Math.min(minOdd, odd[i]);
maxOdd = Math.max(maxOdd, odd[i]);
sumOddArr += odd[i];
}
// To store the total terms in the series
// and the required sum of the array
let totalTerms = 0, reqSum = 0;
// Total terms from 2 to minEven
totalTerms = parseInt(minEven / 2, 10);
// Sum of all even numbers
// from 2 to minEven
let evenSumMin = (totalTerms * (totalTerms + 1));
// Total terms from 2 to maxEven
totalTerms = parseInt(maxEven / 2, 10);
// Sum of all even numbers from
// 2 to maxEven
let evenSumMax = (totalTerms * (totalTerms + 1));
// Required sum for the even array
reqSum = evenSumMax - evenSumMin + minEven;
// Missing even number
document.write("Even = " + (reqSum - sumEvenArr) + "</br>");
// Total terms from 1 to minOdd
totalTerms = parseInt(minOdd / 2, 10) + 1;
// Sum of all odd numbers
// from 1 to minOdd
let oddSumMin = totalTerms * totalTerms;
// Total terms from 1 to maxOdd
totalTerms = parseInt(maxOdd / 2, 10) + 1;
// Sum of all odd numbers
// from 1 to maxOdd
let oddSumMax = totalTerms * totalTerms;
// Required sum for the odd array
reqSum = oddSumMax - oddSumMin + minOdd;
// Missing odd number
document.write("Odd = " + (reqSum - sumOddArr));
}
let even = [ 6, 4, 8, 14, 10 ];
let sizeEven = even.length;
let odd = [ 7, 5, 3, 11, 13 ];
let sizeOdd = odd.length;
findMissingNums(even, sizeEven, odd, sizeOdd);
</script>
Output:
Even = 12
Odd = 9
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