# 查找兩個排序數組的相對補碼 > 原文： [https://www.geeksforgeeks.org/find-relative-complement-two-sorted-arrays/](https://www.geeksforgeeks.org/find-relative-complement-two-sorted-arrays/) 給定兩個大小分別為 m 和 n 的排序數組 arr1 和 arr2。 我們需要找到兩個數組的相對補碼，即 arr1 – arr2，這意味著我們需要找到 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.取兩個指針 i 和 j 分別穿過 arr1 和 arr2。 2.如果 arr1 [i]元素小于 arr2 [j]元素，則打印此元素并增加 i。 3.如果 arr1 元素大于 arr2 [j]元素，則遞增 j。 4.否則增加 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; } ```