用上一个和下一个的乘法替换每个数组元素 · GeeksForGeeks 中文系列教程

# 用上一个和下一个的乘法替换每个数组元素 > 原文： [https://www.geeksforgeeks.org/replace-every-array-element-by-multiplication-of-previous-and-next/](https://www.geeksforgeeks.org/replace-every-array-element-by-multiplication-of-previous-and-next/) 给定一个整数数组，请使用前一个和下一个元素的乘积更新每个元素，但以下例外。 a）第一个元素被替换为`first`和`second`。 b）将`last`元素替换为`last`和倒数第二乘积。 **示例**： ``` Input: arr[] = {2, 3, 4, 5, 6} Output: arr[] = {6, 8, 15, 24, 30} // We get the above output using following // arr[] = {2*3, 2*4, 3*5, 4*6, 5*6} ``` 资料来源：[前 25 个面试问题](https://www.geeksforgeeks.org/top-25-interview-questions/) 一个简单的解决方案是创建一个辅助数组，将给定数组的内容复制到辅助数组。最后遍历辅助数组并使用复制的值更新给定的数组。此解决方案的时间复杂度为`O(n)`，但需要`O(n)`额外空间。一个有效的解决方案可以解决`O(n)`时间和`O(1)`空间中的问题。这个想法是跟踪循环中的前一个元素。以下是此想法的实现。 ## C ``` // C++ program to update every array element with // multiplication of previous and next numbers in array #include<iostream> using namespace std; void modify(int arr[], int n) { // Nothing to do when array size is 1 if (n <= 1) return; // store current value of arr[0] and update it int prev = arr[0]; arr[0] = arr[0] * arr[1]; // Update rest of the array elements for (int i=1; i<n-1; i++) { // Store current value of next interation int curr = arr[i]; // Update current value using previos value arr[i] = prev * arr[i+1]; // Update previous value prev = curr; } // Update last array element arr[n-1] = prev * arr[n-1]; } // Driver program int main() { int arr[] = {2, 3, 4, 5, 6}; int n = sizeof(arr)/sizeof(arr[0]); modify(arr, n); for (int i=0; i<n; i++) cout << arr[i] << " "; return 0; } ``` ## Java ```java // Java program to update every array element with // multiplication of previous and next numbers in array import java.io.*; import java.util.*; import java.lang.Math; class Multipy { static void modify(int arr[], int n) { // Nothing to do when array size is 1 if (n <= 1) return; // store current value of arr[0] and update it int prev = arr[0]; arr[0] = arr[0] * arr[1]; // Update rest of the array elements for (int i=1; i<n-1; i++) { // Store current value of next interation int curr = arr[i]; // Update current value using previos value arr[i] = prev * arr[i+1]; // Update previous value prev = curr; } // Update last array element arr[n-1] = prev * arr[n-1]; } // Driver program to test above function public static void main(String[] args) { int arr[] = {2, 3, 4, 5, 6}; int n = arr.length; modify(arr, n); for (int i=0; i<n; i++) System.out.print(arr[i]+" "); } } /* This code is contributed by Devesh Agrawal */ ``` ## Python3 ```py # Python program to update every array element with # multiplication of previous and next numbers in array def modify(arr, n): # Nothing to do when array size is 1 if n <= 1: return # store current value of arr[0] and update it prev = arr[0] arr[0] = arr[0] * arr[1] # Update rest of the array elements for i in range(1, n-1): # Store current value of next interation curr = arr[i]; # Update current value using previos value arr[i] = prev * arr[i+1] # Update previous value prev = curr # Update last array element arr[n-1] = prev * arr[n-1] # Driver program arr = [2, 3, 4, 5, 6] n = len(arr) modify(arr, n) for i in range (0, n): print(arr[i],end=" ") # This code is contributed by # Smitha Dinesh Semwal ``` ## C# ```cs // C# program to update every array // element with multiplication of // previous and next numbers in array using System; class GFG { static void modify(int []arr, int n) { // Nothing to do when array size is 1 if (n <= 1) return; // store current value of arr[0] and update it int prev = arr[0]; arr[0] = arr[0] * arr[1]; // Update rest of the array elements for (int i=1; i<n-1; i++) { // Store current value of next interation int curr = arr[i]; // Update current value using previos value arr[i] = prev * arr[i+1]; // Update previous value prev = curr; } // Update last array element arr[n-1] = prev * arr[n-1]; } // Driver program to test above function public static void Main() { int []arr = {2, 3, 4, 5, 6}; int n = arr.Length; modify(arr, n); for (int i=0; i<n; i++) Console.Write(arr[i]+" "); } } // This code is contributed by Sam007 ``` ## PHP ```php <?php // PHP program to update every array // element with multiplication of previous // and next numbers in array function modify(&$arr, $n) { // Nothing to do when array size is 1 if ($n <= 1) return; // store current value of arr[0] // and update it $prev = $arr[0]; $arr[0] = $arr[0] * $arr[1]; // Update rest of the array elements for ($i = 1; $i < $n - 1; $i++) { // Store current value of // next interation $curr = $arr[$i]; // Update current value using // previos value $arr[$i] = $prev * $arr[$i + 1]; // Update previous value $prev = $curr; } // Update last array element $arr[$n-1] = $prev * $arr[$n - 1]; } // Driver Code $arr = array (2, 3, 4, 5, 6); $n = sizeof($arr); modify($arr, $n); for ($i = 0; $i < $n; $i++) echo $arr[$i] ." "; // This code is contributed // by ChitraNayal ?> ``` 输出： ``` 6 8 15 24 30 ```