重温数据结构-栈的应用：进制转换，括号匹配检测，行编辑，迷宫求解，求表达式的值

1.括号匹配检测，行编辑，迷宫求解代码

/* * $filename: MyStackApplication.java,v $ * $Date: 2014-3-11  $ * Copyright (C) ZhengHaibo, Inc. All rights reserved. * This software is Made by Zhenghaibo. */package edu.njupt.zhb;import java.util.Stack;/* *@author: ZhengHaibo   *web:     http://blog.csdn.net/nuptboyzhb *mail:    [e-mail] *2014-3-11  Nanjing,njupt,China *//** * 使用栈完成相应的算法 */public class MyStackApplication {/** * @param args */public static void main(String[] args) {// TODO Auto-generated method stub        new MyStackApplication().conversion(100,8);        System.out.println(new MyStackApplication().isMacth("({[]})(){[}[]"));        System.out.println(new MyStackApplication().lineEdit("if(@whli##ilr#e(s#*s)"));        char[][] maze = {{'1','1','1','1','1','1','1','1','1','1'}, {'1','0','0','1','1','1','0','0','1','1'}, {'1','0','0','1','1','0','0','1','0','1'}, {'1','0','0','0','0','0','0','1','0','1'}, {'1','0','0','0','0','1','1','0','0','1'}, {'1','0','0','1','1','1','0','0','0','1'}, {'1','0','0','0','0','1','0','1','0','1'}, {'1','0','1','1','0','0','0','1','0','1'}, {'1','1','0','0','0','0','1','0','0','1'}, {'1','1','1','1','1','1','1','1','1','1'}};      new MyStackApplication().mazeExit(maze,8,8,1,7);        }/** * 将10进制数字n转化为k进制 * @param n * @param k */public void conversion(int n,final int k){Stack stack = new Stack();while(n!=0){int num = n%k;stack.push(num);n=n/k;}while(!stack.isEmpty()){System.out.print(stack.pop());}}/** * 判断字符串中的括号是否匹配 * @param string * @return */public boolean isMacth(String string){char [] str = string.toCharArray();Stack stack = new Stack();for(int i=0;i<str.length;i++){switch (str[i]) {case '{':case '[':case '(':stack.push(Integer.valueOf(str[i]));break;case '}':char c1 = (char)((Integer)stack.pop()).intValue();if(c1!='{'){return false;}break;case ']':char c2 = (char)((Integer)stack.pop()).intValue();if(c2!='['){return false;}break;case ')':char c3 = (char)((Integer)stack.pop()).intValue();if(c3!='('){return false;}break;default:break;}}return true;}/** * 处理行编辑程序 * 解释：当用户发现刚刚输入了一个错误的字符，可补进一个退格“#”， * 用于表示前一个字符无效；如果发现之前输入的难以补救，则插入一个退行符‘@’，表示之前的均无效 * 例如： * 字符串：if(@whli##ilr#e(s#*s) * 实际有效：while(*s) * @param string */public String lineEdit(String string){Stack stack = new Stack();char []str = string.toCharArray();for(int i=0;i<str.length;i++){if(str[i]=='#'){//如果发现一个#，就弹出一个字符if(!stack.isEmpty()){stack.pop();}}else if(str[i]=='@'){//如果发现一个@，就清空栈stack.clear();}else {//将字符加入栈stack.push(Integer.valueOf(str[i]));}}//将stack中的元素放入另一个栈，进行反转Stack result = new Stack();while(!stack.isEmpty()){result.push(stack.pop());}//将result栈中的元素放入到数组中char []resultStr = new char[result.size()];int i=0;while(!result.isEmpty()){resultStr[i++]=(char)((Integer)result.pop()).intValue();}return new String(resultStr);}/** *保存迷宫中的一个单元格  */public class Cell{int x;//单元格所在行int y;//单元格所在列boolean isVisited=false;char c = ' ';//用于保存是墙、可通路或起点到终点的路径public Cell(int x,int y,boolean isVisited,char c){this.x = x;this.y = y;this.isVisited = isVisited;this.c = c;}}/** * 打印迷宫的可通路径 * @param maze 迷宫矩阵 * @param startX 起点横坐标 * @param startY 起点纵坐标 * @param endX 终点横坐标 * @param endY 终点纵坐标 */public void mazeExit(char [][]maze,int startX,int startY,int endX,int endY){Cell [][]cells = createMaze(maze);System.out.println("---------------------");displayMaze(cells);Stack<Cell> stack = new Stack<Cell>();Cell startCell = cells[startX][startY];Cell endCell = cells[endX][endY];startCell.isVisited=true;stack.push(startCell);while(!stack.isEmpty()){Cell currentCell = stack.peek();int x = currentCell.x;int y = currentCell.y;if(currentCell==endCell){//找到了终点while(!stack.isEmpty()){Cell cell = stack.pop();//取出终点cell.c='*';//设置为可通路径//栈中除了含有路径之外，还包含了未继续探索的单元while (!stack.isEmpty()&&!isNearByCell(stack.peek(),cell)){//不连续相邻的，删除stack.pop();}}System.out.println("---------------------");displayMaze(cells);return;}else{//未找到终点之前boolean isContinue = false;if(!cells[x+1][y].isVisited&&cells[x+1][y].c=='0'){//右 (未被访问且不是强)cells[x+1][y].isVisited = true;stack.push(cells[x+1][y]);isContinue = true;}if(!cells[x][y+1].isVisited&&cells[x][y+1].c=='0'){//下 (未被访问且不是强)cells[x][y+1].isVisited = true;stack.push(cells[x][y+1]);isContinue = true;}if(!cells[x-1][y].isVisited&&cells[x-1][y].c=='0'){//左 (未被访问且不是强)cells[x-1][y].isVisited = true;stack.push(cells[x-1][y]);isContinue = true;}if(!cells[x][y-1].isVisited&&cells[x][y-1].c=='0'){//上 (未被访问且不是强)cells[x][y-1].isVisited = true;stack.push(cells[x][y-1]);isContinue = true;}if(!isContinue){//该节点的周围都不能继续访问了，删除之stack.pop();}}}}/** * 判断是否是邻近的单元格 * @param cell1 * @param cell2 * @return */private boolean isNearByCell(Cell cell1, Cell cell2) {// TODO Auto-generated method stubif(cell1.x==cell2.x&&Math.abs(cell1.y-cell2.y)==1){//上下相邻return true;}if(cell1.y==cell2.y&&Math.abs(cell1.x-cell2.x)==1){//左右相邻return true;}return false;}/** * 根据迷宫矩阵，创建迷宫的Cell矩阵 * @param maze * @return */public Cell[][] createMaze(char[][] maze) {// TODO Auto-generated method stubCell [][]cells = new Cell[maze.length][];for(int i = 0;i<maze.length;i++){cells[i] = new Cell[maze[i].length];for(int j=0;j<maze[i].length;j++){cells[i][j]=new Cell(i, j, false, maze[i][j]);}}return cells;}/** * 打印迷宫 * @param cells */public void displayMaze(Cell [][]cells){for(int i=0;i<cells.length;i++){for(int j=0;j<cells[i].length;j++){System.out.print(cells[i][j].c);}System.out.println();}}}

运行结果：

144false
while(*s)
---------------------
1111111111
1001110011
1001100101
1000000101
1000011001
1001110001
1000010101
1011000101
1100001001
1111111111
---------------------
1111111111
100111**11
10011**101
1*****0101
1*00011001
1*0111***1
1****1*1*1
1011***1*1
11000010*1
1111111111

2.求表达式的值：

参考博客：http://blog.csdn.net/txg703003659/article/details/6926100