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:    [email protected]
 *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 stack = new Stack();
		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 stub
		if(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 stub
		Cell [][]cells = new Cell[maze.length][];
		for(int i = 0;i

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