数组中奇数位置的元素增加 1，偶数位置的元素减少 1
原文:https://www . geeksforgeeks . org/在数组中按 1 递增奇数位置元素和按 1 递减偶数位置元素/
给定一个数组 arr[] ，任务是将所有奇数定位元素递增 1 ，将所有偶数定位元素递减 1 。
示例:
输入: arr[] = {3，6，8} 输出: 4 5 9
输入: arr[] = {9，7，3} 输出: 10 6 4
方法:逐元素遍历数组元素，如果当前元素的位置是奇数，则递增 1，否则递减 1。最后打印更新后数组的内容。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Utility function to print
// the contents of an array
void printArr(int arr[], int n)
{
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
}
// Function to increment all the odd
// positioned elements by 1 and decrement
// all the even positioned elements by 1
void updateArr(int arr[], int n)
{
for (int i = 0; i < n; i++)
// If current element is odd positioned
if ((i + 1) % 2 == 1)
arr[i]++;
// If even positioned
else
arr[i]--;
// Print the updated array
printArr(arr, n);
}
// Driver code
int main()
{
int arr[] = { 3, 6, 8 };
int n = sizeof(arr) / sizeof(arr[0]);
updateArr(arr, n);
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
{
// Utility function to print
// the contents of an array
static void printArr(int arr[], int n)
{
for (int i = 0; i < n; i++)
System.out.print(arr[i] + " ");
}
// Function to increment all the odd
// positioned elements by 1 and decrement
// all the even positioned elements by 1
static void updateArr(int arr[], int n)
{
for (int i = 0; i < n; i++)
// If current element is odd positioned
if ((i + 1) % 2 == 1)
arr[i]++;
// If even positioned
else
arr[i]--;
// Print the updated array
printArr(arr, n);
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 3, 6, 8 };
int n = arr.length;
updateArr(arr, n);
}
}
// This code is contributed by Prerna Saini
Python 3
# Python3 implementation of the approach
# Utility function to print
# the contents of an array
def printArr(arr, n):
for i in range(0, n):
print(arr[i], end = " ");
# Function to increment all the odd
# positioned elements by 1 and decrement
# all the even positioned elements by 1
def updateArr(arr, n):
for i in range(0, n):
# If current element is odd positioned
if ((i + 1) % 2 == 1):
arr[i] += 1;
# If even positioned
else:
arr[i] -= 1;
# Print the updated array
printArr(arr, n);
# Driver code
if __name__ == '__main__':
arr = [3, 6, 8];
n = len(arr);
updateArr(arr, n);
# This code contributed by PrinciRaj1992
C
// C# implementation of the approach
class GfG
{
// Utility function to print
// the contents of an array
static void printArr(int []arr, int n)
{
for (int i = 0; i < n; i++)
System.Console.Write(arr[i] + " ");
}
// Function to increment all the odd
// positioned elements by 1 and decrement
// all the even positioned elements by 1
static void updateArr(int []arr, int n)
{
for (int i = 0; i < n; i++)
// If current element is odd positioned
if ((i + 1) % 2 == 1)
arr[i]++;
// If even positioned
else
arr[i]--;
// Print the updated array
printArr(arr, n);
}
// Driver code
static void Main()
{
int []arr = { 3, 6, 8 };
int n = arr.Length;
updateArr(arr, n);
}
}
// This code is contributed by mits
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Utility function to print
// the contents of an array
function printArr($arr, $n)
{
for ($i = 0; $i < $n; $i++)
echo $arr[$i] . " ";
}
// Function to increment all the odd
// positioned elements by 1 and decrement
// all the even positioned elements by 1
function updateArr($arr, $n)
{
for ($i = 0; $i < $n; $i++)
// If current element is odd positioned
if (($i + 1) % 2 == 1)
$arr[$i]++;
// If even positioned
else
$arr[$i]--;
// Print the updated array
printArr($arr, $n);
}
// Driver code
$arr = array( 3, 6, 8 );
$n = count($arr);
updateArr($arr, $n);
// This code is contributed by mits
?>
java 描述语言
<script>
// javascript implementation of the approach
// Utility function to print
// the contents of an array
function printArr(arr, n)
{
var i;
for (i = 0; i < n; i++)
document.write(arr[i] + " ");
}
// Function to increment all the odd
// positioned elements by 1 and decrement
// all the even positioned elements by 1
function updateArr(arr, n)
{
var i;
for (i = 0; i < n; i++)
// If current element is odd positioned
if ((i + 1) % 2 == 1)
arr[i]++;
// If even positioned
else
arr[i]--;
// Print the updated array
printArr(arr, n);
}
// Driver code
var arr = [3, 6, 8];
var n = arr.length;
updateArr(arr, n);
</script>
Output:
4 5 9
时间复杂度: O(n)
辅助空间: O(1)
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