使两个数组的元素相同，最小增减 · GeeksForGeeks 中文系列教程

# 使两个数组的元素相同，最小增减 > 原文： [https://www.geeksforgeeks.org/making-elements-of-two-arrays-same-with-minimum-incrementdecrement/](https://www.geeksforgeeks.org/making-elements-of-two-arrays-same-with-minimum-incrementdecrement/) 给定两个相同大小的数组，我们需要使用最少的操作将第一个数组转换为另一个数组。在一个操作中，我们可以将元素增加或减少一个。请注意，元素的出现顺序不必相同。为了将一个数字转换为另一个数字，我们可以对其加减 1。 **示例**： > **输入**： a = {3，1，1}，b = {1，2，2} > **输出**： 2 > **说明**：在这里，我们可以通过 1 运算将任何 1 变成 2，并以一个减数运算将 3 变成 2。因此 a []变成{2，2，1}，这是 b []的排列。 > > **输入**： a = {3，1，1}，b = {1，1，2} > **输出**： 1 **算法**： 1.首先对两个数组进行排序。 2.排序后，我们将运行一个循环，在该循环中，我们将比较第一个和第二个数组元素，并计算使第一个数组等于第二个数组所需的操作。下面是上述方法的实现 ## C++ ```cpp // CPP program to find minimum increment/decrement // operations to make array elements same. #include <bits/stdc++.h> using namespace std; int MinOperation(int a[], int b[], int n) { // sorting both arrays in // ascending order sort(a, a + n); sort(b, b + n); // variable to store the // final result int result = 0; // After sorting both arrays // Now each array is in non- // decreasing order. Thus, // we will now compare each // element of the array and // do the increment or decrement // operation depending upon the // value of array b[]. for (int i = 0; i < n; ++i) { result = result + abs(a[i] - b[i]); } return result; } // Driver code int main() { int a[] = { 3, 1, 1 }; int b[] = { 1, 2, 2 }; int n = sizeof(a) / sizeof(a[0]); cout << MinOperation(a, b, n); return 0; } ``` ## Java ```java // Java program to find minimum // increment/decrement operations // to make array elements same. import java.util.Arrays; import java.io.*; class GFG { static int MinOperation(int a[], int b[], int n) { // sorting both arrays // in ascending order Arrays.sort(a); Arrays.sort(b); // variable to store // the final result int result = 0; // After sorting both arrays // Now each array is in non- // decreasing order. Thus, // we will now compare each // element of the array and // do the increment or decrement // operation depending upon the // value of array b[]. for (int i = 0; i < n; ++i) { if (a[i] > b[i]) result = result + Math.abs(a[i] - b[i]); else if (a[i] < b[i]) result = result + Math.abs(a[i] - b[i]); } return result; } // Driver code public static void main (String[] args) { int a[] = {3, 1, 1}; int b[] = {1, 2, 2}; int n = a.length; System.out.println(MinOperation(a, b, n)); } } // This code is contributed // by akt_mit ``` ## Python3 ```py # Python 3 program to find minimum # increment/decrement operations to # make array elements same. def MinOperation(a, b, n): # sorting both arrays in ascending order a.sort(reverse = False) b.sort(reverse = False) # variable to store the final result result = 0 # After sorting both arrays. Now each # array is in non-decreasing order. # Thus, we will now compare each element # of the array and do the increment or # decrement operation depending upon # the value of array b[]. for i in range(0, n, 1): if (a[i] > b[i]): result = result + abs(a[i] - b[i]) elif(a[i] < b[i]): result = result + abs(a[i] - b[i]) return result # Driver code if __name__ == '__main__': a = [3, 1, 1] b = [1, 2, 2] n = len(a) print(MinOperation(a, b, n)) # This code is contributed by # Sahil_Shelangia ``` ## C# ```cs //C# program to find minimum // increment/decrement operations // to make array elements same. using System; public class GFG { static int MinOperation(int []a, int []b, int n) { // sorting both arrays // in ascending order Array.Sort(a); Array.Sort(b); // variable to store // the final result int result = 0; // After sorting both arrays // Now each array is in non- // decreasing order. Thus, // we will now compare each // element of the array and // do the increment or decrement // operation depending upon the // value of array b[]. for (int i = 0; i < n; ++i) { if (a[i] > b[i]) result = result + Math.Abs(a[i] - b[i]); else if (a[i] < b[i]) result = result + Math.Abs(a[i] - b[i]); } return result; } // Driver code public static void Main () { int []a = {3, 1, 1}; int []b = {1, 2, 2}; int n = a.Length; Console.WriteLine(MinOperation(a, b, n)); } } /*This C# code is contributed by 29AjayKumar*/ ``` ## PHP ```php <?php // PHP program to find minimum // increment/decrement operations // to make array elements same. function MinOperation($a, $b, $n) { // sorting both arrays in // ascending order sort($a); sort($b); // variable to store // the final result $result = 0; // After sorting both arrays // Now each array is in non- // decreasing order. Thus, // we will now compare each // element of the array and // do the increment or decrement // operation depending upon the // value of array b[]. for ($i = 0; $i < $n; ++$i) { if ($a[$i] > $b[$i]) $result = $result + abs($a[$i] - $b[$i]); else if ($a[$i] < $b[$i]) $result = $result + abs($a[$i] - $b[$i]); } return $result; } // Driver code $a = array ( 3, 1, 1 ); $b = array ( 1, 2, 2 ); $n = sizeof($a); echo MinOperation($a, $b, $n); // This code is contributed by ajit ?> ``` **输出**： ``` 2 ``` **时间复杂度**：`O(N log N)` [](https://practice.geeksforgeeks.org/courses/competitive-programming-live?utm_source=geeksforgeeks&utm_medium=article&utm_campaign=gfg_article_cp) * * * * * *