您的位置:首页 > 编程语言 > Java开发

JAVA经典题目汇总

2011-02-28 15:38 381 查看
转 :http://bea.javaeye.com/blog/166344

1 Fibonacci

Fibonacci为1200年代的欧洲数学家,在他的著作中曾经提到:若有一只兔子每个月生一只小兔子,一个月后小兔子也开始生产。起初只有一只兔子,一个月后就有两只兔子,两个月后有三只兔子,三个月后有五只兔子(小兔子投入生产)……
这就是Fibonacci数列,一般习惯称之为费式数列,例如:1,1,2,3,5,8,13,21,34,55,89,……

java代码:

public class Fibonacci {
public static void main(String[] args) {
int[] fib = new int[20];
fib[0] = 0;
fib[1] = 1;
for(int i = 2; i < fib.length; i++)
fib[i] = fib[i-1] + fib[i-2];
for(int i = 0; i < fib.length; i++)
System.out.print(fib[i] + " ");
System.out.println();
}
}


2 巴斯卡(Pascal)

三角形基本上就是在解nCr ,因为三角形上的每一个数字各对应一个nCr ,其中n为row,而r为colnmu

java代码:

import java.awt.*;
import javax.swing.*;
public class Pascal extends JFrame {
public Pascal() {
setBackground(Color.white);
setTitle("巴斯卡三角形");
setSize(520, 350);
setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
show();
}
private long combi(int n, int r){
int i;
long p = 1;
for(i = 1; i <= r; i++)
p = p * (n-i+1) / i;

return p;
}
public void paint(Graphics g) {
final int N = 12;
int n, r, t;
for(n = 0; n <= N; n++) {
for(r = 0; r <= n; r++)
g.drawString(" " + combi(n, r),
(N-n)*20 + r * 40, n * 20 + 50);
}
}
public static void main(String args[]) {
Pascal frm = new Pascal();
}
}


3 ThreeColorFlags

ThreeColorFlags问题最早由E.W.Dijkstra所提出,塔所使用的用语为Dutch Nation Flag(Dijkstra为荷兰人),而多数的作者则使用Three-Color Flag来说明。

假设有一条绳子,上面有红,白,蓝三种颜色的旗子,起初绳子上的旗子颜色并没有顺序,您希望将之分类,并排列蓝,白,红的顺序,要如何移动次数才会最少,注意您只能在绳子上进行这个动作,而且一次只能调换两个旗子。

java代码:

import java.io.*;
public class ThreeColorsFlags {
private void swap(char[] flags, int x, int y) {
char temp;
temp = flags[x];
flags[x] = flags[y];
flags[y] = temp;
}

public String move(char[] flags) {
int wFlag = 0;
int bFlag = 0;
int rFlag = flags.length - 1;

while(wFlag <= rFlag) {
if(flags[wFlag] == 'W') {
wFlag++;
}
else if(flags[wFlag] == 'B') {
swap(flags, bFlag, wFlag);
bFlag++;
wFlag++;
}
else {
while(wFlag < rFlag && flags[rFlag] == 'R')
rFlag--;
swap(flags, rFlag, wFlag);
rFlag--;
}
}

return new String(flags);
}

public static void main(String[] args)
throws IOException {
BufferedReader buf;
buf = new BufferedReader(
new InputStreamReader(System.in));
System.out.print("输入三色旗顺序(ex. RWBBWRWR):");
String flags = buf.readLine();

ThreeColorsFlags threeColorsFlag = new ThreeColorsFlags();
flags = threeColorsFlag.move(
flags.toUpperCase().toCharArray());

System.out.println("移动顺序后:" + flags);
}
}


4 Mouse

Mouse走迷宫是循环求解的基本类型,我们在二维数组中用2来表示迷宫的墙壁,使用1来表示老鼠的行走路径,并用程序求出从入口到出口的距离。

java代码:

public class Mouse {
private int startI, startJ;  // 入口
private int endI, endJ;  // 出口
private boolean success = false;

public static void main(String[] args) {
int[][] maze = {{2, 2, 2, 2, 2, 2, 2},
{2, 0, 0, 0, 0, 0, 2},
{2, 0, 2, 0, 2, 0, 2},
{2, 0, 0, 2, 0, 2, 2},
{2, 2, 0, 2, 0, 2, 2},
{2, 0, 0, 0, 0, 0, 2},
{2, 2, 2, 2, 2, 2, 2}};

System.out.println("显示迷宫:");
for(int i = 0; i < maze.length; i++) {
for(int j = 0; j < maze[0].length; j++)
if(maze[j] == 2)
System.out.print("█");
else
System.out.print("  ");
System.out.println();
}
Mouse mouse = new Mouse();
mouse.setStart(1, 1);
mouse.setEnd(5, 5);

if(!mouse.go(maze)) {
System.out.println("/n没有找到出口!");
}
else {
System.out.println("/n找到出口!");
for(int i = 0; i < maze.length; i++) {
for(int j = 0; j < maze[0].length; j++) {
if(maze[j] == 2)
System.out.print("█");
else if(maze[j] == 1)
System.out.print("◇");
else
System.out.print("  ");
}
System.out.println();
}
}
}

public void setStart(int i, int j) {
this.startI = i;
this.startJ = j;
}

public void setEnd(int i, int j) {
this.endI = i;
this.endJ = j;
}

public boolean go(int[][] maze) {
return visit(maze, startI, startJ);
}

private boolean visit(int[][] maze, int i, int j) {
maze[j] = 1;
if(i == endI && j == endJ)
success = true;
if(!success && maze[j+1] == 0)
visit(maze, i, j+1);
if(!success && maze[i+1][j] == 0)
visit(maze, i+1, j);
if(!success && maze[j-1] == 0)
visit(maze, i, j-1);
if(!success && maze[i-1][j] == 0)
visit(maze, i-1, j);
if(!success)
maze[j] = 0;

return success;
}
}


5 Knight tour

骑士游戏,在十八世纪倍受数学家与拼图迷的注意,骑士的走法为西洋棋的走发,骑士可以由任何一个位置出发,它要如何走完所有的位置。

java代码:

public class Knight {
public boolean travel(int startX,
int startY, int[][] board) {
// 对应骑士可以走的八个方向
int[] ktmove1 = {-2, -1, 1, 2, 2, 1, -1, -2};
int[] ktmove2 = {1, 2, 2, 1, -1, -2, -2, -1};

// 下一个出路的位置         int[] nexti = new int[board.length];
int[] nextj = new int[board.length];

// 记录出路的个数
int[] exists = new int[board.length];

int x = startX;
int y = startY;

board[x][y] = 1;

for(int m = 2; m <= Math.pow(board.length, 2); m++) {
for(int k = 0; k < board.length; k++) {
exists[k] = 0;
}

int count = 0;
// 试探八个方向
for(int k = 0; k < board.length; k++) {
int tmpi = x + ktmove1[k];
int tmpj = y + ktmove2[k];

// 如果是边界,不可以走
if(tmpi < 0 || tmpj < 0 ||
tmpi > 7 || tmpj > 7) {
continue;
}

// 如果这个方向可以走,记录下来
if(board[tmpi][tmpj] == 0) {
nexti[count] = tmpi;
nextj[count] = tmpj;
// 可走的方向加一个
count++;
}
}

int min = -1;
if(count == 0) {
return false;
}
else if(count == 1) {
min = 0;
}
else {
// 找出下个位置的出路数
for(int l = 0; l < count; l++) {
for(int k = 0; k < board.length; k++) {
int tmpi = nexti[l] + ktmove1[k];
int tmpj = nextj[l] + ktmove2[k];
if(tmpi < 0 || tmpj < 0 ||
tmpi > 7 || tmpj > 7) {
continue;
}
if(board[tmpi][tmpj] == 0)
exists[l]++;
}
}
int tmp = exists[0];
min = 0;
// 从可走的方向寻找最少出路的方向
for(int l = 1; l < count; l++) {
if(exists[l] < tmp) {
tmp = exists[l];
min = l;
}
}
}

// 走最少出路的方向
x = nexti[min];
y = nextj[min];
board[x][y] = m;
}

return true;
}

public static void main(String[] args) {
int[][] board = new int[8][8];
Knight knight = new Knight();

if(knight.travel(
Integer.parseInt(args[0]),
Integer.parseInt(args[1]), board)) {
System.out.println("走棋完成!");
}
else {
System.out.println("走棋失败!");
}

for(int i = 0; i < board.length; i++) {
for(int j = 0; j < board[0].length; j++) {
if(board[i][j] < 10) {
System.out.print(" " + board[i][j]);
}
else {
System.out.print(board[i][j]);
}
System.out.print(" ");
}
System.out.println();
}
}
}


6 Queen

西洋棋中的皇后可以直线前进,吃掉遇到的所有棋子,如果棋盘上有八个皇后,则这八个皇后如何相安无事的放置在棋盘上?

java代码:

public class Queen {
// 同位置是否有皇后,1表示有
private int[] column;
// 右上至左下是否有皇后
private int[] rup;
// 左上至右下是否有皇后
private int[] lup;
// 解答
private int[] queen;

// 解答编号
private int num;

public Queen() {
column = new int[8+1];
rup = new int[2*8+1];
lup = new int[2*8+1];

for(int i = 1; i <= 8; i++)
column[i] = 1;
for(int i = 1; i <= 2*8; i++)
rup[i] = lup[i] = 1;

queen = new int[8+1];
}

public void backtrack(int i) {
if(i ><IMG src="/images/smiles/icon_cool.gif" mce_src="images/smiles/icon_cool.gif">{
showAnswer();
}
else {
for(int j = 1; j <= 8; j++) {
if(column[j] == 1 &&
rup[i+j] == 1 &&
lup[i-j+8] == 1) {
queen[i] = j;
// 设定为占用
column[j] = rup[i+j] = lup[i-j+8] = 0;
backtrack(i+1);
column[j] = rup[i+j] = lup[i-j+8] = 1;
}
}
}
}

protected void showAnswer() {
num++;
System.out.println("/n解答 " + num);
for(int y = 1; y <= 8; y++) {
for(int x = 1; x <= 8; x++) {
if(queen[y] == x) {
System.out.print(" Q");
}
else {
System.out.print(" .");
}
}
System.out.println();
}
}

public static void main(String[] args) {
Queen queen = new Queen();
queen.backtrack(1);
}
}


7 Coins

现在有八枚银币abcdefg,已知其中一枚是假币,其重量不同于真币,但不知道是轻还是重,如何用天平以最小的比较次数决定出那个是假币,并得知假币是比真币轻还是重。

java代码:

public class Coins {
private int[] coins;

public Coins() {
coins = new int[8];
for(int i = 0; i < 8; i++)
coins[i] = 10;
}

public void setFake(int weight) {
coins[(int) (Math.random() * 7)] = weight;
}

public void fake() {
if(coins[0]+coins[1]+coins[2] ==
coins[3]+coins[4]+coins[5]) {
if(coins[6] > coins[7])
compare(6, 7, 0);
else
compare(7, 6, 0);
}
else if(coins[0]+coins[1]+coins[2] >
coins[3]+coins[4]+coins[5]) {
if(coins[0]+coins[3] == coins[1]+coins[4])
compare(2, 5, 0);
else if(coins[0]+coins[3] > coins[1]+coins[4])
compare(0, 4, 1);
if(coins[0]+coins[3] < coins[1]+coins[4])
compare(1, 3, 0);
}
else if(coins[0]+coins[1]+coins[2] <
coins[3]+coins[4]+coins[5]) {
if(coins[0]+coins[3] == coins[1]+coins[4])
compare(5, 2, 0);
else if(coins[0]+coins[3] > coins[1]+coins[4])
compare(3, 1, 0);
if(coins[0]+coins[3] < coins[1]+coins[4])
compare(4, 0, 1);
}
}

protected void compare(int i, int j, int k) {
if(coins[i] > coins[k])
System.out.print("/n假币 " + (i+1) + " 较重");
else
System.out.print("/n假币 " + (j+1) + " 较轻");
}

public static void main(String[] args) {
if(args.length == 0) {
System.out.println("输入假币重量(比10大或小)");
System.out.println("ex. java Coins 5");
return;
}

Coins eightCoins = new Coins();
eightCoins.setFake(Integer.parseInt(args[0]));
eightCoins.fake();
}
}


8 Life game

生命游戏,为1970年英国数学家J.H.Conway所提出,某一细胞的邻居包括上,下,左,右,左上,左下,右上与右下相邻的细胞,游戏规则如下:

1,孤单死亡:如果细胞的邻居小于一个,则该细胞在下一个状态死亡。

2,拥挤死亡:如果细胞的邻居在四个以上,则该细胞在下一个状态死亡。

3,稳定:如果细胞的邻居为两个或三个,则该细胞在下一个状态稳定。

4,复活:如果某位置原无细胞存活,而该位置的邻居为三个,则该位置将复活一个细胞。

java代码:

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class LifeGame {
private boolean[][] map;
private boolean[][] newmap;

public LifeGame(int maxRow, int maxColumn) {
map = new boolean[maxRow][maxColumn];
newmap = new boolean[maxRow][maxColumn];
}

public void setCell(int x, int y) {
map[x][y] = true;
}

public void next() {
for(int row = 0; row < map.length; row++) {
for(int col = 0; col < map[0].length; col++) {
switch (neighbors(row, col)) {
case 0:
case 1:
case 4:
case 5:
case 6:
case 7:
case 8:
newmap[row][col] = false;
break;
case 2:
newmap[row][col] = map[row][col];
break;
case 3:
newmap[row][col] = true;
break;
}
}
}
copyMap();
}
public void outputMap() throws IOException {
System.out.println("/n/nGame of life cell status");
for(int row = 0; row < map.length; row++) {
System.out.print("/n ");
for(int col = 0; col < map[0].length; col++)
if(map[row][col] == true)
System.out.print('#');
else
System.out.print('-');
}
}

private void copyMap() {
for(int row = 0; row < map.length; row++)
for(int col = 0; col < map[0].length; col++)
map[row][col] = newmap[row][col];
}

private int neighbors(int row, int col) {
int count = 0;
for(int r = row-1; r <= row+1; r++)
for(int c = col-1; c <= col+1; c++) {
if(r < 0 || r >= map.length ||
c < 0 || c >= map[0].length)
continue;
if(map[r][c] == true)
count++;
}
if(map[row][col] == true)
count--;

return count;
}

public static void main(String[] args)
throws NumberFormatException, IOException {
BufferedReader bufReader =
new BufferedReader(
new InputStreamReader(System.in));

LifeGame game = new LifeGame(10, 25);

System.out.println("Game of life Program");
System.out.println(
"Enter x, y where x, y is living cell");
System.out.println("0 <= x < 10, 0 <= y < 25");
System.out.println("Terminate with x, y = -1, -1");

while(true) {
String[] strs = bufReader.readLine().split(" ");
int row = Integer.parseInt(strs[0]);
int col = Integer.parseInt(strs[1]);
if(0 <= row && row < 10 && 0 <= col && row < 25)
game.setCell(row, col);
else if(row == -1 || col == -1) {
break;
}
else {
System.out.print(
"(x, y) exceeds map ranage!");
}
}

while(true) {
game.outputMap();
game.next();
System.out.print(
"/nContinue next Generation ? ");

String ans = bufReader.readLine().toUpperCase();
if(!ans.equals("Y"))
break;
}
}
}


9 String Match

现在的一些高级程序语言对于字符串的处理支持越来越大,不过字符串搜寻本身仍是值得探讨的课题,在这里以Boyer Moore法来说明如何进行字符串说明,这个方法速度快且容易理解。

java代码:

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class StringMatch {
private int[] skip;
private int p;
private String str;
private String key;

public StringMatch(String key) {
skip = new int[256];
this.key = key;

for(int k = 0; k <= 255; k++)
skip[k] = key.length();
for(int k = 0; k < key.length() - 1; k++)
skip[key.charAt(k)] = key.length() - k - 1;
}

public void search(String str) {
this.str = str;
p = search(key.length()-1, str, key);
}

private int search(int p, String input, String key) {
while(p < input.length()) {
String tmp = input.substring(
p-key.length()+1, p+1);
if(tmp.equals(key))  // 比较两个字符串是否相同
return p-key.length()+1;
p += skip[input.charAt(p)];
}
return -1;
}

public boolean hasNext() {
return (p != -1);
}

public String next() {
String tmp = str.substring(p);
p = search(p+key.length()+1, str, key);
return tmp;
}

public static void main(String[] args)
throws IOException {
BufferedReader bufReader =
new BufferedReader(
new InputStreamReader(System.in));

System.out.print("请输入字符串:");
String str = bufReader.readLine();
System.out.print("请输入搜寻关键字:");
String key = bufReader.readLine();

StringMatch strMatch = new StringMatch(key);
strMatch.search(str);
while(strMatch.hasNext()) {
System.out.println(strMatch.next());
}
}
}


10 Kanpsack Problem

假设一个背包的负重最大可达8公斤,而希望在背包内放置负重范围你价值最高的物品。

java代码:

class Fruit {
private String name;
private int size;
private int price;

public Fruit(String name, int size, int price) {
this.name = name;
this.size = size;
this.price = price;
}

public String getName() {
return name;
}
public int getPrice() {
return price;
}
public int getSize() {
return size;
}
}
public class Knapsack {
public static void main(String[] args) {
final int MAX = 8;
final int MIN = 1;
int[] item = new int[MAX+1];
int[] value = new int[MAX+1];
Fruit fruits[] = {
new Fruit("李子", 4, 4500),
new Fruit("苹果", 5, 5700),
new Fruit("桔子", 2, 2250),
new Fruit("草莓", 1, 1100),
new Fruit("甜瓜", 6, 6700)};
for(int i = 0; i < fruits.length; i++) {
for(int s = fruits[i].getSize(); s <= MAX; s++) {
int p = s - fruits[i].getSize();
int newvalue = value[p] +
fruits[i].getPrice();
if(newvalue > value[s]) {// 找到阶段最佳解
value[s] = newvalue;
item[s] = i;
}
}
}
System.out.println("物品/t价格");
for(int i = MAX;
i >= MIN;
i = i - fruits[item[i]].getSize()) {
System.out.println(fruits[item[i]].getName()+
"/t" + fruits[item[i]].getPrice());
}
System.out.println("合计/t" + value[MAX]);
}
}


11 Towers of Hanoi

河內之塔(Towers of Hanoi)是法国人M.Claus(Lucas)於1883年从泰国带至法国的,河內为越战时北越的首都,即现在的胡志明市;1883年法国数学家 Edouard Lucas曾提及這个故事,据说创世紀时Benares有一座波罗教塔,是由三支钻石棒(Pag)所支撑,开始时神在第一根棒上放置64个由上至下依由小至大排列的金盘(Disc),並命令僧侣将所有的金盘从第一根石棒移至第三根石棒,且搬运过程中遵守大盘子在小盘子之下的原则,若每日仅搬一个盘子,则当盘子全数搬运完毕之时,此塔将损毁,而也就是世界末日來临之时。

java代码:

import java.io.*;
public class Hanoi {
public static void main(String args[]) throws IOException {
int n;
BufferedReader buf;
buf = new BufferedReader(new InputStreamReader(System.in));
System.out.print("请输入盘子个数");
n = Integer.parseInt(buf.readLine());
Hanoi hanoi = new Hanoi();
hanoi.move(n, 'A', 'B', 'C');
}
public void move(int n, char a, char b, char c) {
if(n == 1)
System.out.println("盘 " + n + " 由 " + a + " 移至 " + c);
else {
move(n - 1, a, c, b);
System.out.println("盘 " + n + " 由 " + a + " 移至 " + c);
move(n - 1, b, a, c);
}
}
}


12 Hanoi2Colors

Hanoi2Colors是由河内塔演变而来的一种算法。

java代码:

public class Hanoi2Colors {
public static void help() {
System.out.println(
"Usage: java Hanoi2Colors number_of_disks");
System.out.println(
"/t number_of_disks: must be a even number.");
System.exit(0);
}

public static void main(String[] args) {
int disks = 0;
try {
disks = Integer.parseInt(args[0]);
} catch (Exception e) {
help();
}
if ((disks <= 0) || (disks % 2 != 0)) {
help();
}
hanoi2colors(disks);
}

public static void hanoi(int disks,
char source, char temp, char target) {
if (disks == 1) {
System.out.println("move disk from "
+ source + " to " + target);
System.out.println("move disk from "
+ source + " to " + target);
} else {
hanoi(disks-1, source, target, temp);
hanoi(1, source, temp, target);
hanoi(disks-1, temp, source, target);
}
}
public static void hanoi2colors(int disks) {
char source = 'A';
char temp = 'B';
char target = 'C';
for (int i = disks / 2; i > 1; i--) {
hanoi(i-1, source, temp, target);
System.out.println("move disk from "
+ source + " to " + temp);
System.out.println("move disk from "
+ source + " to " + temp);
hanoi(i-1, target, temp, source);
System.out.println("move disk from "
+ temp + " to " + target);
}
System.out.println("move disk from "
+ source + " to " + temp);
System.out.println("move disk from "
+ source + " to " + target);
}
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: