# 將第一個元素加倍，然后將零移動到結尾 > 原文： [https://www.geeksforgeeks.org/double-first-element-move-zero-end/](https://www.geeksforgeeks.org/double-first-element-move-zero-end/) 給定大小為 **n** 的整數數組。 假設“`0`”為無效數字，所有其他均為有效數字。 轉換數組的方式是，如果下一個數字是有效數字并且與當前數字相同，則將其值加倍，然后用 0 替換下一個數字。修改后，重新排列數組，使所有 0 都移到末尾。 例子： ``` Input : arr[] = {2, 2, 0, 4, 0, 8} Output : 4 4 8 0 0 0 Input : arr[] = {0, 2, 2, 2, 0, 6, 6, 0, 0, 8} Output : 4 2 12 8 0 0 0 0 0 0 ``` **來源**： [Microsoft IDC 面試體驗| 系列 150](https://www.geeksforgeeks.org/microsoft-idc-interview-experience-set-150-full-time/) **方法**：首先修改提及的數組，即，如果下一個有效數字與當前數字相同，則將其值加倍，然后將下一個數字替換為 0。 **修改算法**： ``` 1\. if n == 1 2\. return 3\. for i = 0 to n-2 4\. if (arr[i] != 0) && (arr[i] == arr[i+1]) 5\. arr[i] = 2 * arr[i] 6\. arr[i+1] = 0 7\. i++ ``` 修改數組后，[將所有零移動到數組](https://www.geeksforgeeks.org/move-zeroes-end-array/)的末尾。 ## C++ ```cpp // C++ implementation to rearrange the array? // elements after modification #include <bits/stdc++.h> using namespace std; // function which pushes all zeros to end of? // an array. void pushZerosToEnd(int arr[], int n) { ????// Count of non-zero elements ????int count = 0; ????// Traverse the array. If element encountered ????// is non-zero, then replace the element at? ????// index 'count' with this element ????for (int i = 0; i < n; i++) ????????if (arr[i] != 0) ????????????// here count is incremented ????????????arr[count++] = arr[i]; ????// Now all non-zero elements have been shifted ????// to front and 'count' is set as index of ????// first 0\. Make all elements 0 from count ????// to end. ????while (count < n) ????????arr[count++] = 0; } // function to rearrange the array elements // after modification void modifyAndRearrangeArr(int arr[], int n) { ????// if 'arr[]' contains a single element ????// only ????if (n == 1) ????????return; ????// traverse the array ????for (int i = 0; i < n - 1; i++) { ????????// if true, perform the required modification ????????if ((arr[i] != 0) && (arr[i] == arr[i + 1])) { ????????????// double current index value ????????????arr[i] = 2 * arr[i]; ????????????// put 0 in the next index ????????????arr[i + 1] = 0; ????????????// increment by 1 so as to move two? ????????????// indexes ahead during loop iteration ????????????i++; ????????} ????} ????// push all the zeros at the end of 'arr[]' ????pushZerosToEnd(arr, n); } // function to print the array elements void printArray(int arr[], int n) { ????for (int i = 0; i < n; i++) ????????cout << arr[i] << " "; } // Driver program to test above int main() { ????int arr[] = { 0, 2, 2, 2, 0, 6, 6, 0, 0, 8 }; ????int n = sizeof(arr) / sizeof(arr[0]); ????cout << "Original array: "; ????printArray(arr, n); ????modifyAndRearrangeArr(arr, n); ????cout << "\nModified array: "; ????printArray(arr, n); ????return 0; } ``` ## Java ```java // Java implementation to rearrange the? // array elements after modification class GFG { ????// function which pushes all? ????// zeros to end of an array. ????static void pushZerosToEnd(int arr[], int n) ????{ ????????// Count of non-zero elements ????????int count = 0; ????????// Traverse the array. If element? ????????// encountered is non-zero, then ????????// replace the element at index ????????// 'count' with this element ????????for (int i = 0; i < n; i++) ????????????if (arr[i] != 0) ????????????????// here count is incremented ????????????????arr[count++] = arr[i]; ????????// Now all non-zero elements? ????????// have been shifted to front and? ????????// 'count' is set as index of first 0.? ????????// Make all elements 0 from count to end. ????????while (count < n) ????????????arr[count++] = 0; ????} ????// function to rearrange the array ????//? elements after modification ????static void modifyAndRearrangeArr(int arr[], int n) ????{ ????????// if 'arr[]' contains a single element ????????// only ????????if (n == 1) ????????????return; ????????// traverse the array ????????for (int i = 0; i < n - 1; i++) { ????????????// if true, perform the required modification ????????????if ((arr[i] != 0) && (arr[i] == arr[i + 1])) ????????????{ ????????????????// double current index value ????????????????arr[i] = 2 * arr[i]; ????????????????// put 0 in the next index ????????????????arr[i + 1] = 0; ????????????????// increment by 1 so as to move two ????????????????// indexes ahead during loop iteration ????????????????i++; ????????????} ????????} ????????// push all the zeros at? ????????// the end of 'arr[]' ????????pushZerosToEnd(arr, n); ????} ????// function to print the array elements ????static void printArray(int arr[], int n) ????{ ????????for (int i = 0; i < n; i++) ????????????System.out.print(arr[i] + " "); ????????System.out.println(); ????} ????// Driver program to test above ????public static void main(String[] args) ????{ ????????int arr[] = { 0, 2, 2, 2, 0, 6, 6, 0, 0, 8 }; ????????int n = arr.length; ????????System.out.print("Original array: "); ????????printArray(arr, n); ????????modifyAndRearrangeArr(arr, n); ????????System.out.print("Modified array: "); ????????printArray(arr, n); ????} } // This code is contributed?? // by prerna saini ``` ## Python3 ```py # Python3 implementation to rearrange? # the array elements after modification # function which pushes all zeros? # to end of an array. def pushZerosToEnd(arr, n): ????# Count of non-zero elements ????count = 0 ????# Traverse the array. If element? ????# encountered is non-zero, then? ????# replace the element at index? ????# 'count' with this element ????for i in range(0, n): ????????if arr[i] != 0: ????????????# here count is incremented ????????????arr[count] = arr[i] ????????????count+=1 ????# Now all non-zero elements have been? ????# shifted to front and 'count' is set ????# as index of first 0\. Make all? ????# elements 0 from count to end. ????while (count < n): ????????arr[count] = 0 ????????count+=1 # function to rearrange the array # elements after modification def modifyAndRearrangeArr(ar, n): ????# if 'arr[]' contains a single ????# element only ????if n == 1: ????????return ????# traverse the array ????for i in range(0, n - 1): ????????# if true, perform the required modification ????????if (arr[i] != 0) and (arr[i] == arr[i + 1]): ????????????# double current index value ????????????arr[i] = 2 * arr[i] ????????????# put 0 in the next index ????????????arr[i + 1] = 0 ????????????# increment by 1 so as to move two? ????????????# indexes ahead during loop iteration ????????????i+=1 ????# push all the zeros at the end of 'arr[]' ????pushZerosToEnd(arr, n) # function to print the array elements def printArray(arr, n): ????for i in range(0, n): ????????print(arr[i],end=" ") # Driver program to test above arr = [ 0, 2, 2, 2, 0, 6, 6, 0, 0, 8 ] n = len(arr)? print("Original array:",end=" ") printArray(arr, n) modifyAndRearrangeArr(arr, n) print("\nModified array:",end=" ") printArray(arr, n) # This code is contributed by Smitha Dinesh Semwal ``` ## C# ```cs // C# implementation to rearrange the // array elements after modification using System; class GFG { ????// function which pushes all ????// zeros to end of an array. ????static void pushZerosToEnd(int[] arr, int n) ????{ ????????// Count of non-zero elements ????????int count = 0; ????????// Traverse the array. If element ????????// encountered is non-zero, then ????????// replace the element at index ????????// 'count' with this element ????????for (int i = 0; i < n; i++) ????????????if (arr[i] != 0) ????????????????// here count is incremented ????????????????arr[count++] = arr[i]; ????????// Now all non-zero elements ????????// have been shifted to front and ????????// 'count' is set as index of first 0\. ????????// Make all elements 0 from count to end. ????????while (count < n) ????????????arr[count++] = 0; ????} ????// function to rearrange the array ????// elements after modification ????static void modifyAndRearrangeArr(int[] arr, int n) ????{ ????????// if 'arr[]' contains a single element ????????// only ????????if (n == 1) ????????????return; ????????// traverse the array ????????for (int i = 0; i < n - 1; i++) { ????????????// if true, perform the required modification ????????????if ((arr[i] != 0) && (arr[i] == arr[i + 1])) { ????????????????// double current index value ????????????????arr[i] = 2 * arr[i]; ????????????????// put 0 in the next index ????????????????arr[i + 1] = 0; ????????????????// increment by 1 so as to move two ????????????????// indexes ahead during loop iteration ????????????????i++; ????????????} ????????} ????????// push all the zeros at ????????// the end of 'arr[]' ????????pushZerosToEnd(arr, n); ????} ????// function to print the array elements ????static void printArray(int[] arr, int n) ????{ ????????for (int i = 0; i < n; i++) ????????????Console.Write(arr[i] + " "); ????????Console.WriteLine(); ????} ????// Driver program to test above ????public static void Main() ????{ ????????int[] arr = { 0, 2, 2, 2, 0, 6, 6, 0, 0, 8 }; ????????int n = arr.Length; ????????Console.Write("Original array: "); ????????printArray(arr, n); ????????modifyAndRearrangeArr(arr, n); ????????Console.Write("Modified array: "); ????????printArray(arr, n); ????} } // This code is contributed by Sam007 ``` Output: ``` Original array: 0 2 2 2 0 6 6 0 0 8 Modified array: 4 2 12 8 0 0 0 0 0 0 ``` 時間復雜度：`O(n)`。 * * * * * *