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]))
例子：
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.从数组A，B和C中最大的元素开始。在要执行的每个步骤中，都维护一个变量以更新答案。
在每个步骤中，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answer变量。
重复此步骤，直到包含最大元素的数组结束。
C++
// C++ code for above approach
#include
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. Python3
# python code for above approach.
def solve(A, B, C):
# assigning the length -1 value
# to each of three variables
i = len(A) - 1
j = len(B) - 1
k = len(C) - 1
# calculating min difference
# from last index of lists
min_diff = abs(max(A[i], B[j], C[k]) -
min(A[i], B[j], C[k]))
while i != -1 and j != -1 and k != -1:
current_diff = abs(max(A[i], B[j],
C[k]) - min(A[i], 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], B[j], C[k])
# Moving to smaller value in the
# array with maximum out of three.
if A[i] == max_term:
i -= 1
elif B[j] == max_term:
j -= 1
else:
k -= 1
return min_diff
# driver code
A = [ 5, 8, 10, 15 ]
B = [ 6, 9, 15, 78, 89 ]
C = [ 2, 3, 6, 6, 8, 8, 10 ]
print(solve(A, B, C))C#
// C# code for above approach
using System;
class GFG
{
static int solve(int[] A, int[] B, int[] C)
{
int i, j, k;
// assigning the length -1 value
// to each of three variables
i = A.Length - 1;
j = B.Length - 1;
k = C.Length - 1;
int min_diff, current_diff, max_term;
// calculating min difference
// from last index of lists
min_diff = Math.Abs(Math.Max(A[i], Math.Max(B[j], C[k]))
- Math.Min(A[i], Math.Min(B[j], C[k])));
while (i != -1 && j != -1 && k != -1)
{
current_diff = Math.Abs(Math.Max(A[i], Math.Max(B[j], C[k]))
- Math.Min(A[i], Math.Min(B[j], C[k])));
// checking condition
if (current_diff < min_diff)
min_diff = current_diff;
// calculating max term from list
max_term = Math.Max(A[i], Math.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
public static void Main()
{
int[] D = { 5, 8, 10, 15 };
int[] E = { 6, 9, 15, 78, 89 };
int[] F = { 2, 3, 6, 6, 8, 8, 10 };
Console.WriteLine(solve(D, E, F));
}
}
// This code is contributed by vt_mPHP
Javascript
输出：
1