# 最小化三個不同排序數組的（max（A [i]，B [j]，C [k]）– min（A [i]，B [j]，C [k]）） > 原文： [https://www.geeksforgeeks.org/minimize-maxai-bj-ck-minai-bj-ck-three-different-sorted-arrays/](https://www.geeksforgeeks.org/minimize-maxai-bj-ck-minai-bj-ck-three-different-sorted-arrays/) 給定三個排序數組 A，B 和 C，它們的大小不一定相同。 計算任何三元組 A [i]，B [j]，C [k]的最大數目和最小數目之間的最小絕對差，以使它們分別屬于數組 A，B 和 C，即最小化（max（A [i ]，B [j]，C [k]）– min（A [i]，B [j]，C [k]）） **示例**： ``` Input : A : [ 1, 4, 5, 8, 10 ] B : [ 6, 9, 15 ] C : [ 2, 3, 6, 6 ] Output : 1 Explanation: When we select A[i] = 5 B[j] = 6, C[k] = 6, we get the minimum difference as max(A[i], B[j], C[k]) - min(A[i], B[j], C[k])) = |6-5| = 1 Input : A = [ 5, 8, 10, 15 ] B = [ 6, 9, 15, 78, 89 ] C = [ 2, 3, 6, 6, 8, 8, 10 ] Output : 1 Explanation: When we select A[i] = 10 b[j] = 9, C[k] = 10. ``` 從每個數組 A，B & C 中最大的元素開始。在隨后的每個步驟中，都維護一個變量以更新答案。 在每一步中，減小差異的唯一可能方法是減小三個元素中的最大元素。 因此遍歷包含此步驟中最大元素的數組中的下一個最大元素，并更新答案變量。 重復此步驟，直到包含最大元素的數組結束。 ## C++ ```cpp // C++ code for above approach #include<bits/stdc++.h> using namespace std; int solve(int A[], int B[], int C[], int i, int j, int k) {? ????????int min_diff, current_diff, max_term; ????????// calculating min difference from last ????????// index of lists ????????min_diff = Integer.MAX_VALUE; ????????while (i != -1 && j != -1 && k != -1)? ????????{ ????????????current_diff = abs(max(A[i], max(B[j], C[k]))? ????????????????????????????- min(A[i], min(B[j], C[k]))); ????????????// checking condition ????????????if (current_diff < min_diff) ????????????????min_diff = current_diff; ????????????// calculating max term from list ????????????max_term = max(A[i], max(B[j], C[k])); ????????????// Moving to smaller value in the ????????????// array with maximum out of three. ????????????if (A[i] == max_term) ????????????????i -= 1; ????????????else if (B[j] == max_term) ????????????????j -= 1; ????????????else ????????????????k -= 1; ????????} ????????return min_diff; ????} ????// Driver code ????int main() ????{ ????????int D[] = { 5, 8, 10, 15 }; ????????int E[] = { 6, 9, 15, 78, 89 }; ????????int F[] = { 2, 3, 6, 6, 8, 8, 10 }; ????????int nD = sizeof(D) / sizeof(D[0]); ????????int nE = sizeof(E) / sizeof(E[0]); ????????int nF = sizeof(F) / sizeof(F[0]); ????????cout << solve(D, E, F, nD-1, nE-1, nF-1); ????????return 0;? ????} // This code is contributed by Ravi Maurya. ```