從其對和數組構造一個數組 · GeeksForGeeks 中文系列教程

# 從其對和數組構造一個數組 > 原文： [https://www.geeksforgeeks.org/construct-array-pair-sum-array/](https://www.geeksforgeeks.org/construct-array-pair-sum-array/) 給定一個成對和數組和原始數組的大小（n），構造原始數組。數組的成對和數組是包含有序形式的所有對的和的數組。例如，arr [] = {6，8，3，4}的成對和數組為{14，9，10，11，12，7}。通常，arr [0..n-1]的成對和數組為{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}。 “給出一個成對和數組，構造原始數組。” **我們強烈建議您最小化瀏覽器并自己嘗試。** 假設給定的數組為“ pair []”，并且原始數組中有 n 個元素。如果我們看幾個示例，我們可以看到 arr [0]是 pair [0] + pair [1] – pair [n-1]的一半。請注意，對[0] +對[1] –對[n-1]的值是（arr [0] + arr [1]）+（arr [0] + arr [2]）–（arr [1 ] + arr [2]）。一旦我們評估了 arr [0]，我們就可以通過減去 arr [0]來評估其他元素。例如，可以通過從對[0]中減去 arr [0]來評估 arr [1]，通過從對[1]中減去 arr [0]來評估 arr [2]。以下是上述想法的實現。 ## 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 ?> ``` **輸出**： ``` 8 7 5 3 2 ``` ConstructArr（）的時間復雜度為`O(n)`，其中 n 是 arr []中元素的數量。本文由 Abhishek 提供。如果發現任何不正確的地方，或者想分享有關上述主題的更多信息，請寫評論。