从其对和数组构造一个数组 · GeeksForGeeks 中文系列教程

# 从其对和数组构造一个数组 > 原文： [https://www.geeksforgeeks.org/construct-array-pair-sum-array/](https://www.geeksforgeeks.org/construct-array-pair-sum-array/) 给定一个成对和数组和原始数组的大小（n），构造原始数组。数组的成对和数组是包含有序形式的所有对的和的数组。例如，arr [] = {6，8，3，4}的成对和数组为{14，9，10，11，12，7}。通常，arr [0..n-1]的成对和数组为{arr [0] + arr [1]，arr [0] + arr [2]，……。，arr [1] + arr [ 2]，arr [1] + arr [3]，……。，arr [2] + arr [3]，arr [2] + arr [4]，…。，arr [n-2] + arr [n -1}。 “给出一个成对和数组，构造原始数组。” **我们强烈建议您最小化浏览器并自己尝试。** 假设给定的数组为“ pair []”，并且原始数组中有 n 个元素。如果我们看几个示例，我们可以看到 arr [0]是 pair [0] + pair [1] – pair [n-1]的一半。请注意，对[0] +对[1] –对[n-1]的值是（arr [0] + arr [1]）+（arr [0] + arr [2]）–（arr [1 ] + arr [2]）。一旦我们评估了 arr [0]，我们就可以通过减去 arr [0]来评估其他元素。例如，可以通过从对[0]中减去 arr [0]来评估 arr [1]，通过从对[1]中减去 arr [0]来评估 arr [2]。以下是上述想法的实现。 ## C++ ```cpp #include <bits/stdc++.h> using namespace std; // Fills element in arr[] from its pair sum array pair[]. // n is size of arr[] void constructArr(int arr[], int pair[], int n) { arr[0] = (pair[0]+pair[1]-pair[n-1]) / 2; for (int i=1; i<n; i++) arr[i] = pair[i-1]-arr[0]; } // Driver program to test above function int main() { int pair[] = {15, 13, 11, 10, 12, 10, 9, 8, 7, 5}; int n = 5; int arr[n]; constructArr(arr, pair, n); for (int i = 0; i < n; i++) cout << arr[i] << " "; return 0; } ``` ## Java ```java import java.io.*; class PairSum { // Fills element in arr[] from its pair sum array pair[]. // n is size of arr[] static void constructArr(int arr[], int pair[], int n) { arr[0] = (pair[0]+pair[1]-pair[n-1]) / 2; for (int i=1; i<n; i++) arr[i] = pair[i-1]-arr[0]; } // Driver program to test above function public static void main(String[] args) { int pair[] = {15, 13, 11, 10, 12, 10, 9, 8, 7, 5}; int n = 5; int[] arr = new int[n]; constructArr(arr, pair, n); for (int i = 0; i < n; i++) System.out.print(arr[i]+" "); } } /* This code is contributed by Devesh Agrawal */ ``` ## Python3 ```py # Fills element in arr[] from its # pair sum array pair[]. # n is size of arr[] def constructArr(arr,pair,n): arr [0] = (pair[0]+pair[1]-pair[n-1])//2 for i in range(1,n): arr[i] = pair[i-1]-arr[0] # Driver code if __name__=='__main__': pair = [15, 13, 11, 10, 12, 10, 9, 8, 7, 5] n =5 arr = [0]*n constructArr(arr,pair,n) for i in range(n): print(arr[i],end =" ") # This code is contributed by # Shrikant13 ``` ## C# ```cs // C# program to construct an // array from its pair-sum array using System; class PairSum { // Fills element in arr[] from its // pair sum array pair[]. // n is size of arr[] static void constructArr(int []arr, int []pair, int n) { arr[0] = (pair[0] + pair[1] - pair[n - 1]) / 2; for (int i = 1; i < n; i++) arr[i] = pair[i - 1] - arr[0]; } // Driver program to test above function public static void Main() { int []pair = {15, 13, 11, 10, 12, 10, 9, 8, 7, 5}; int n = 5; int []arr = new int[n]; constructArr(arr, pair, n); for (int i = 0; i < n; i++) Console.Write(arr[i] + " "); } } // This code is contributed by nitin mittal ``` ## PHP ```php <?php // Fills element in arr[] from // its pair sum array pair[]. // n is size of arr[] function constructArr($pair) { $arr = array(); $n = 5; $arr[0] = intval(($pair[0] + $pair[1] - $pair[$n - 1]) / 2); for ($i = 1; $i < $n; $i++) $arr[$i] = $pair[$i - 1] - $arr[0]; for ($i = 0; $i < $n; $i++) echo $arr[$i] . " "; } // Driver Code $pair = array(15, 13, 11, 10, 12, 10, 9, 8, 7, 5); constructArr($pair); // This code is contributed by Sam007 ?> ``` **输出**： ``` 8 7 5 3 2 ``` ConstructArr（）的时间复杂度为`O(n)`，其中 n 是 arr []中元素的数量。本文由 Abhishek 提供。如果发现任何不正确的地方，或者想分享有关上述主题的更多信息，请写评论。