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;

}

}