# 查找两个排序数组的相对补码 > 原文： [https://www.geeksforgeeks.org/find-relative-complement-two-sorted-arrays/](https://www.geeksforgeeks.org/find-relative-complement-two-sorted-arrays/) 给定两个大小分别为 m 和 n 的排序数组 arr1 和 arr2。 我们需要找到两个数组的相对补码，即 arr1 – arr2，这意味着我们需要找到 arr1 中存在但 arr2 中不存在的所有元素。 例子： ``` Input : arr1[] = {3, 6, 10, 12, 15} arr2[] = {1, 3, 5, 10, 16} Output : 6 12 15 The elements 6, 12 and 15 are present in arr[], but not present in arr2[] Input : arr1[] = {10, 20, 36, 59} arr2[] = {5, 10, 15, 59} Output : 20 36 ``` 1.取两个指针 i 和 j 分别穿过 arr1 和 arr2。 2.如果 arr1 [i]元素小于 arr2 [j]元素，则打印此元素并增加 i。 3.如果 arr1 元素大于 arr2 [j]元素，则递增 j。 4.否则增加 i 和 j。 ## C++ ```cpp // CPP program to find all those  // elements of arr1[] that are not // present in arr2[] #include <iostream> using namespace std; void relativeComplement(int arr1[], int arr2[],                                int n, int m) {   int i = 0, j = 0;   while (i < n && j < m) {     // If current element in arr2[] is     // greater, then arr1[i] can't be      // present in arr2[j..m-1]     if (arr1[i] < arr2[j]) {       cout << arr1[i] << " ";       i++;     // Skipping smaller elements of     // arr2[]     } else if (arr1[i] > arr2[j]) {       j++;     // Equal elements found (skipping     // in both arrays)     } else if (arr1[i] == arr2[j]) {       i++;       j++;     }   }   // Printing remaining elements of   // arr1[]   while (i < n)      cout << arr1[i] << " ";   } // Driver code int main() {   int arr1[] = {3, 6, 10, 12, 15};   int arr2[] = {1, 3, 5, 10, 16};   int n = sizeof(arr1) / sizeof(arr1[0]);   int m = sizeof(arr2) / sizeof(arr2[0]);   relativeComplement(arr1, arr2, n, m);   return 0; } ```