s= a*Math.pow(10,i)*(n-i);结果是double类型。。s是int类型。。double比int大。。所以可能损失精度。。改为 s= (int) (a*Math.pow(10,i)*(n-i))；或者将s改为double。
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代码实现(Java)
1. 输入
(1) 代表地图二值二维数组(0表示可通路，1表示路障)
int[][] maps = {
{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },
{ 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },
{ 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0 },
{ 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0 },
{ 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0 },
{ 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0 },
{ 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0 }
};123456789123456789
(2) 按照二维数组的特点，坐标原点在左上角，所以y是高，x是宽，y向下递增，x向右递增，我们将x和y封装成一个类，好传参，重写equals方法比较坐标(x,y)是不是同一个。
public class Coord
{
public int x;
public int y;
public Coord(int x, int y)
{
this.x = x;
this.y = y;
}
@Override
public boolean equals(Object obj)
{
if (obj == null) return false;
if (obj instanceof Coord)
{
Coord c = (Coord) obj;
return x == c.x y == c.y;
}
return false;
}
}12345678910111213141516171819202122231234567891011121314151617181920212223
(3) 封装路径结点类，字段包括：坐标、G值、F值、父结点，实现Comparable接口，方便优先队列排序。
public class Node implements Comparable
{
public Coord coord; // 坐标
public Node parent; // 父结点
public int G; // G：是个准确的值，是起点到当前结点的代价
public int H; // H：是个估值，当前结点到目的结点的估计代价
public Node(int x, int y)
{
this.coord = new Coord(x, y);
}
public Node(Coord coord, Node parent, int g, int h)
{
this.coord = coord;
this.parent = parent;
G = g;
H = h;
}
@Override
public int compareTo(Node o)
{
if (o == null) return -1;
if (G + H o.G + o.H)
return 1;
else if (G + H o.G + o.H) return -1;
return 0;
}
}1234567891011121314151617181920212223242526272829303112345678910111213141516171819202122232425262728293031
(4) 最后一个数据结构是A星算法输入的所有数据，封装在一起，传参方便。:grin:
public class MapInfo
{
public int[][] maps; // 二维数组的地图
public int width; // 地图的宽
public int hight; // 地图的高
public Node start; // 起始结点
public Node end; // 最终结点
public MapInfo(int[][] maps, int width, int hight, Node start, Node end)
{
this.maps = maps;
this.width = width;
this.hight = hight;
this.start = start;
this.end = end;
}
}12345678910111213141516171234567891011121314151617
2. 处理
(1) 在算法里需要定义几个常量来确定：二维数组中哪个值表示障碍物、二维数组中绘制路径的代表值、计算G值需要的横纵移动代价和斜移动代价。
public final static int BAR = 1; // 障碍值
public final static int PATH = 2; // 路径
public final static int DIRECT_VALUE = 10; // 横竖移动代价
public final static int OBLIQUE_VALUE = 14; // 斜移动代价12341234
(2) 定义两个辅助表：Open表和Close表。Open表的使用是需要取最小值，在这里我们使用Java工具包中的优先队列PriorityQueue，Close只是用来保存结点，没其他特殊用途，就用ArrayList。
Queue openList = new PriorityQueue(); // 优先队列(升序)
List closeList = new ArrayList();1212
(3) 定义几个布尔判断方法：最终结点的判断、结点能否加入open表的判断、结点是否在Close表中的判断。
/**
* 判断结点是否是最终结点
*/
private boolean isEndNode(Coord end,Coord coord)
{
return coord != null end.equals(coord);
}
/**
* 判断结点能否放入Open列表
*/
private boolean canAddNodeToOpen(MapInfo mapInfo,int x, int y)
{
// 是否在地图中
if (x 0 || x = mapInfo.width || y 0 || y = mapInfo.hight) return false;
// 判断是否是不可通过的结点
if (mapInfo.maps[y][x] == BAR) return false;
// 判断结点是否存在close表
if (isCoordInClose(x, y)) return false;
return true;
}
/**
* 判断坐标是否在close表中
*/
private boolean isCoordInClose(Coord coord)
{
return coord!=nullisCoordInClose(coord.x, coord.y);
}
/**
* 判断坐标是否在close表中
*/
private boolean isCoordInClose(int x, int y)
{
if (closeList.isEmpty()) return false;
for (Node node : closeList)
{
if (node.coord.x == x node.coord.y == y)
{
return true;
}
}
return false;
}1234567891011121314151617181920212223242526272829303132333435363738394041424344454612345678910111213141516171819202122232425262728293031323334353637383940414243444546
(4) 计算H值，“曼哈顿” 法，坐标分别取差值相加
private int calcH(Coord end,Coord coord)
{
return Math.abs(end.x - coord.x) + Math.abs(end.y - coord.y);
}12341234
(5) 从Open列表中查找结点
private Node findNodeInOpen(Coord coord)
{
if (coord == null || openList.isEmpty()) return null;
for (Node node : openList)
{
if (node.coord.equals(coord))
{
return node;
}
}
return null;
}123456789101112123456789101112
(6) 添加邻结点到Open表
/**
* 添加所有邻结点到open表
*/
private void addNeighborNodeInOpen(MapInfo mapInfo,Node current)
{
int x = current.coord.x;
int y = current.coord.y;
// 左
addNeighborNodeInOpen(mapInfo,current, x - 1, y, DIRECT_VALUE);
// 上
addNeighborNodeInOpen(mapInfo,current, x, y - 1, DIRECT_VALUE);
// 右
addNeighborNodeInOpen(mapInfo,current, x + 1, y, DIRECT_VALUE);
// 下
addNeighborNodeInOpen(mapInfo,current, x, y + 1, DIRECT_VALUE);
// 左上
addNeighborNodeInOpen(mapInfo,current, x - 1, y - 1, OBLIQUE_VALUE);
// 右上
addNeighborNodeInOpen(mapInfo,current, x + 1, y - 1, OBLIQUE_VALUE);
// 右下
addNeighborNodeInOpen(mapInfo,current, x + 1, y + 1, OBLIQUE_VALUE);
// 左下
addNeighborNodeInOpen(mapInfo,current, x - 1, y + 1, OBLIQUE_VALUE);
}
/**
* 添加一个邻结点到open表
*/
private void addNeighborNodeInOpen(MapInfo mapInfo,Node current, int x, int y, int value)
{
if (canAddNodeToOpen(mapInfo,x, y))
{
Node end=mapInfo.end;
Coord coord = new Coord(x, y);
int G = current.G + value; // 计算邻结点的G值
Node child = findNodeInOpen(coord);
if (child == null)
{
int H=calcH(end.coord,coord); // 计算H值
if(isEndNode(end.coord,coord))
{
child=end;
child.parent=current;
child.G=G;
child.H=H;
}
else
{
child = new Node(coord, current, G, H);
}
openList.add(child);
}
else if (child.G G)
{
child.G = G;
child.parent = current;
// 重新调整堆
openList.add(child);
}
}
}1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606112345678910111213141516171819202122232425262728293031323334353637383940414243444546474849505152535455565758596061
(7) 回溯法绘制路径
private void drawPath(int[][] maps, Node end)
{
if(end==null||maps==null) return;
System.out.println("总代价：" + end.G);
while (end != null)
{
Coord c = end.coord;
maps[c.y][c.x] = PATH;
end = end.parent;
}
}12345678910111234567891011
(8) 开始算法，循环移动结点寻找路径，设定循环结束条件，Open表为空或者最终结点在Close表
public void start(MapInfo mapInfo)
{
if(mapInfo==null) return;
// clean
openList.clear();
closeList.clear();
// 开始搜索
openList.add(mapInfo.start);
moveNodes(mapInfo);
}
/**
* 移动当前结点
*/
private void moveNodes(MapInfo mapInfo)
{
while (!openList.isEmpty())
{
if (isCoordInClose(mapInfo.end.coord))
{
drawPath(mapInfo.maps, mapInfo.end);
break;
}
Node current = openList.poll();
closeList.add(current);
addNeighborNodeInOpen(mapInfo,current);
}
}
单元和区域和数值，，，中的最大
import java.util.Scanner;
public class MaxData {
public static void main(String[] args) {
System.out.print("第一个数：");
Scanner scanner1 = new Scanner(System.in);
int input1 = scanner1.nextInt();
System.out.print("第二个数：");
Scanner scanner2 = new Scanner(System.in);
int input2 = scanner2.nextInt();
if (input1 input2) {
System.out.println("第一个数" + input1 + "大");
} else if (input1 input2) {
System.out.println("第二个数" + input2 + "大");
} else {
System.out.println("两个数" + input1 + "和" + input2 + "一样大");
}
scanner1.close();
scanner2.close();
}
}
您好。上面是我写的代码，以及代码运行结果的截图。麻烦看一下是否可以满足要求。
ps： 我的答案只给识货的人看
package zhidao;
import java.util.Arrays;
import java.util.LinkedList;
public class RecursionSubNSort
{
public static void main ( String[] args )
{
int a = 0;
String[] A = { "01", "09", "11", "07", "05", "02" };
String[] B = { "03", "08", "11", "07", "06" };
String bstr = " " + Arrays.toString (B).replaceAll ("[\\[\\],\\s]", ",") + " ";
LinkedListString[] list = new LinkedListString[] ();
recursionSub (list, 1, A, 0, -1);
for ( String[] strings : list )
{
String s = Arrays.toString (strings);
String str = s.replaceAll ("[\\[\\]]", ",");
int i = bstr.split (str).length - 1;
a += i;
System.out.println (s + " 出现在B中的次数有：" + i);
}
System.out.println ("结果 a = " + a);
System.out.println ("======================================");
list.clear ();
a = 0;
recursionSub (list, 2, A, 0, -1);
for ( String[] strings : list )
{
String s = Arrays.toString (strings);
System.out.println ("组合 " + s);
for ( String string : strings )
{
int i = bstr.split ("," + string + ",").length - 1;
a += i;
System.out.println (string + " 在B中出现的次数有：" + i);
}
}
System.out.println ("结果 a = " + a);
System.out.println ("======================================");
list.clear ();
a = 0;
recursionSub (list, 5, A, 0, -1);
for ( String[] strings : list )
{
String s = Arrays.toString (strings);
System.out.println ("组合 " + s);
for ( String string : strings )
{
int i = bstr.split ("," + string + ",").length - 1;
a += i;
System.out.println (string + " 在B中出现的次数有：" + i);
}
}
System.out.println ("结果 a = " + a);
}
private static LinkedListString[] recursionSub ( LinkedListString[] list, int count, String[] array, int ind,
int start, int... indexs )
{
start++;
if (start count - 1)
{
return null;
}
if (start == 0)
{
indexs = new int[array.length];
}
for ( indexs = ind; indexs array.length; indexs++ )
{
recursionSub (list, count, array, indexs + 1, start, indexs);
if (start == count - 1)
{
String[] temp = new String[count];
for ( int i = count - 1; i = 0; i-- )
{
temp[start - i] = array[indexs[start - i]];
}
list.add (temp);
}
}
return list;
}
}
