# 将大小为 n 的数组合并为大小为 m + n 的另一个数组 > 原文： [https://www.geeksforgeeks.org/merge-one-array-of-size-n-into-another-one-of-size-mn/](https://www.geeksforgeeks.org/merge-one-array-of-size-n-into-another-one-of-size-mn/) 有两个排序的数组。 第一个大小为 m + n，仅包含 m 个元素。 另一个大小为 n，包含 n 个元素。 将这两个数组合并到大小为 m + n 的第一个数组中，以便对输出进行排序。 输入：具有 m + n 个元素的数组（mPlusN []）。  NA = >数组 mPlusN []中的值未填充/不可用。 应该有 n 个这样的数组块。 输入：具有 n 个元素的数组（N []）。  输出：N []合并为 mPlusN []（已修改的 mPlusN []）  **算法**： ``` Let first array be mPlusN[] and other array be N[] 1) Move m elements of mPlusN[] to end. 2) Start from nth element of mPlusN[] and 0th element of N[] and merge them into mPlusN[]. ``` 下面是上述算法的实现： ## C++ ```cpp // C++ program to Merge an array of  // size n into another array of size m + n #include <bits/stdc++.h> using namespace std; /* Assuming -1 is filled for the places    where element is not available */ #define NA -1 /* Function to move m elements at     the end of array mPlusN[] */ void moveToEnd(int mPlusN[], int size) {    int j = size - 1;    for (int i = size - 1; i >= 0; i--)      if (mPlusN[i] != NA)      {         mPlusN[j] = mPlusN[i];         j--;      } } /* Merges array N[] of size n into    array mPlusN[] of size m+n*/ int merge(int mPlusN[], int N[], int m, int n) {    int i = n; /* Current index of i/p part of mPlusN[]*/    int j = 0; /* Current index of N[]*/    int k = 0; /* Current index of output mPlusN[]*/    while (k < (m + n))    {     /* Take an element from mPlusN[] if     a) value of the picked element is smaller         and we have not reached end of it     b) We have reached end of N[] */     if ((i < (m + n) && mPlusN[i] <= N[j]) || (j == n))     {         mPlusN[k] = mPlusN[i];         k++;         i++;     }     else // Otherwise take element from N[]     {        mPlusN[k] = N[j];        k++;        j++;     }    } } /* Utility that prints out an array on a line */ void printArray(int arr[], int size) {    for (int i = 0; i < size; i++)    cout << arr[i] << " ";    cout << endl; } /* Driver function to test above functions */ int main() {    /* Initialize arrays */    int mPlusN[] = {2, 8, NA, NA, NA, 13, NA, 15, 20};    int N[] = {5, 7, 9, 25};    int n = sizeof(N) / sizeof(N[0]);    int m = sizeof(mPlusN) / sizeof(mPlusN[0]) - n;    /*Move the m elements at the end of mPlusN*/    moveToEnd(mPlusN, m + n);    /*Merge N[] into mPlusN[] */    merge(mPlusN, N, m, n);    /* Print the resultant mPlusN */    printArray(mPlusN, m+n);    return 0; } ```