# 最小化三个不同排序数组的（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/) 给定三个排序数组 A，B 和 C，它们的大小不一定相同。 计算任何三元组 A [i]，B [j]，C [k]的最大数目和最小数目之间的最小绝对差，以使它们分别属于数组 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. ``` 从每个数组 A，B & C 中最大的元素开始。在随后的每个步骤中，都维护一个变量以更新答案。 在每一步中，减小差异的唯一可能方法是减小三个元素中的最大元素。 因此遍历包含此步骤中最大元素的数组中的下一个最大元素，并更新答案变量。 重复此步骤，直到包含最大元素的数组结束。 ## 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. ```