求一比较算法，接受挑战者入！！

Question

已知下面数组
       i             j           k            i             j            k
     6941.10      2019.27      6630.50      7433.81      1957.73      6648.59 
     7350.81      1328.57      6688.23      6875.53      1371.42      6668.96 
     6941.10      2019.27      6671.98      7433.81      1957.73      6694.96 
     7350.81      1328.57      6732.24      6875.53      1371.42      6708.77 
这个数组组成了一个立方体，每三个值组成一个点的三维坐标，现在求这个立方体的边界值，即长、宽、高的最小值和最大值。

Answer 1

就是分别求出i,j,k上的最大值和最小值

Answer 2

public class MaxMinLengthWidthHeight { public static void main(String[] args) { //所有顶点的三维坐标
double[][] vertixDimensions=
{
{6941.10,2019.27,6630.50},
{7350.81,1328.57,6688.23},
{6941.10,2019.27,6671.98},
{7350.81,1328.57,6732.24},
{7433.81,1957.73,6648.59},
{6875.53,1371.42,6668.96},
{7433.81,1957.73,6694.96},
{6875.53,1371.42,6708.77}

};

//任意取一个定点，不妨就取第一个
double[] source=vertixDimensions[0];

//长宽高中三者最小者的平方
double minSquare=99999999;

//长宽高中三者次小者的平方
double secondMinSquare=99999999;

//某个定点到最远顶点的距离的平方
double maxDistanceSquare=0;

//对每个定点求它到给定点的距离的平方
for(int i=1;i<vertixDimensions.length;i++)
{
double[] target=vertixDimensions[i];
double distanceFromSourceToTargetSquare=
(source[0]-target[0])*
(source[0]-target[0])+
(source[1]-target[1])*
(source[1]-target[1])+
(source[2]-target[2])*
(source[2]-target[2]);

//如果比最小值小
if(minSquare>distanceFromSourceToTargetSquare)
{
//次小值等于最小值
secondMinSquare=minSquare;
//最小值等于求出值
minSquare=distanceFromSourceToTargetSquare;
}
//如果比最小值大，比次小值小
else if(secondMinSquare>distanceFromSourceToTargetSquare)
{
//次小值等于求出值
secondMinSquare=distanceFromSourceToTargetSquare;
}

//如果比最大值大
if(maxDistanceSquare<distanceFromSourceToTargetSquare)
{
//最大值等于求出值
maxDistanceSquare=distanceFromSourceToTargetSquare;
}
}

//输出长宽高
System.out.println("Length:"+Math.sqrt(maxDistanceSquare-minSquare-secondMinSquare));
System.out.println("Width:"+Math.sqrt(secondMinSquare));
System.out.println("Height:"+Math.sqrt(minSquare));
}}