给定对和数组和原始数组的大小(n),构造原始数组。 数组的pair sum数组是包含所有有序对之和的数组。例如,arr[]={6,8,3,4}的对和数组是{14,9,10,11,12,7}。 一般来说,arr[0..n-1]的对和数组是{arr[0]+arr[1]、arr[0]+arr[2]、…、arr[1]+arr[2]、arr[1]+arr[3]、…、arr[2]+arr[4]、…、arr[n-2]+arr[n-1}。 给定对和数组,构造原始数组 我们强烈建议您尽量减少浏览器数量,并亲自尝试。 假设给定的数组是“pair[],并且在原始数组中有n个元素。如果我们看几个例子,我们可以观察到arr[0]是对[0]+对[1]-对[n-1]的一半。请注意,对[0]+对[1]–对[n-1]的值是(arr[0]+对[1])+(arr[0]+对[2])–(arr[1]+对[2])。 一旦我们计算了arr[0],我们就可以通过减去arr[0]来计算其他元素。例如,可以通过从对[0]中减去arr[0]来计算arr[1],也可以通过从对[1]中减去arr[0]来计算arr[2]。 以下是上述理念的实施。
null
C++
#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
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
# 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#
// 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 // 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 ?> |
Javascript
<script> // Fills element in arr[] from // its pair sum array pair[]. // n is size of arr[] function constructArr(arr, pair, n) { arr[0] = Math.floor((pair[0]+pair[1]-pair[n-1]) / 2); for (let i=1; i<n; i++) arr[i] = pair[i-1]-arr[0]; } // Driver program to test above function let pair = [15, 13, 11, 10, 12, 10, 9, 8, 7, 5]; let n = 5; let arr = new Array(n); constructArr(arr, pair, n); for (let i = 0; i < n; i++) document.write(arr[i] + " " ); // This code is contributed by Surbhi Tyagi. </script> |
输出:
8 7 5 3 2
constructArr()的时间复杂度为O(n),其中n是arr[]中的元素数。 本文由阿披实撰稿。如果您发现任何不正确的地方,或者您想分享有关上述主题的更多信息,请写下评论。
© 版权声明
文章版权归作者所有,未经允许请勿转载。
THE END