用java实现a算法代码 java编程计算a+aa+aaa+aaan个a
A*算法java实现
代码实现(Java)
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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
怎么用java实现apriori算法
作者：何史提
链接：
来源：知乎
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
Apriori算法的理念其实很简单，可是实现起上来却复杂无比，因为当中无可避免用Set和Hash Table等高阶的数据结构，而且有很多loop用以读取数据。
我不建议用Java，应改用Python或Scala一类的语言。如果用Python，代码大概50行左右，但可以想像用Java便看起来复杂得多。看如下：
from operator import and_
from itertools import combinations
class AprioriAssociationRule:
def __init__(self, inputfile):
self.transactions = []
self.itemSet = set([])
inf = open(inputfile, 'rb')
for line in inf.readlines():
elements = set(filter(lambda entry: len(entry)0, line.strip().split(',')))
if len(elements)0:
self.transactions.append(elements)
for element in elements:
self.itemSet.add(element)
inf.close()
self.toRetItems = {}
self.associationRules = []
def getSupport(self, itemcomb):
if type(itemcomb) != frozenset:
itemcomb = frozenset([itemcomb])
within_transaction = lambda transaction: reduce(and_, [(item in transaction) for item in itemcomb])
count = len(filter(within_transaction, self.transactions))
return float(count)/float(len(self.transactions))
def runApriori(self, minSupport=0.15, minConfidence=0.6):
itemCombSupports = filter(lambda freqpair: freqpair[1]=minSupport,
map(lambda item: (frozenset([item]), self.getSupport(item)), self.itemSet))
currentLset = set(map(lambda freqpair: freqpair[0], itemCombSupports))
k = 2
while len(currentLset)0:
currentCset = set([i.union(j) for i in currentLset for j in currentLset if len(i.union(j))==k])
currentItemCombSupports = filter(lambda freqpair: freqpair[1]=minSupport,
map(lambda item: (item, self.getSupport(item)), currentCset))
currentLset = set(map(lambda freqpair: freqpair[0], currentItemCombSupports))
itemCombSupports.extend(currentItemCombSupports)
k += 1
for key, supportVal in itemCombSupports:
self.toRetItems[key] = supportVal
self.calculateAssociationRules(minConfidence=minConfidence)
def calculateAssociationRules(self, minConfidence=0.6):
for key in self.toRetItems:
subsets = [frozenset(item) for k in range(1, len(key)) for item in combinations(key, k)]
for subset in subsets:
confidence = self.toRetItems[key] / self.toRetItems[subset]
if confidence minConfidence:
self.associationRules.append([subset, key-subset, confidence])
用JAVA写一个a,b,c,d,e排列组合算法,谢谢了
public class Paixu {
public static void main(String[] args) {
char[] in = "abcde".toCharArray();
new Paixu().paixu(in, in.length, 0);
}
private void paixu(char[] array, int n, int k) {
if (n == k) {
char[] out = new char[n];
for (int i = 0; i array.length; i++) {
out[i] = array[i];
}
System.out.println(new String(out));
} else {
for (int i = k; i n; i++) {
swap(array, k, i);
paixu(array, n, k + 1);
swap(array, i, k);
}
}
}
private void swap(char[] a, int x, int y) {
char temp = a[x];
a[x] = a[y];
a[y] = temp;
}
}
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