算法 练习1.3

1.3.1

import java.io.IOException;

import edu.princeton.cs.algs4.StdOut;


public class FixedCapacityStackOfStrings {
    private String[] a;
    private int N;
    public FixedCapacityStackOfStrings(int cap){
        a = new String[cap];
    }
    public boolean isEmpty(){
        return N == 0;
    }
    public int size(){
        return N;
    }
    public void push(String item){
        a[N++] = item;
    }
    public String pop(){
        return a[--N];
    }
    public boolean isFull(){
        return N == a.length;
    }

    public static void main(String[] args) throws IOException{
        FixedCapacityStackOfStrings  s = new FixedCapacityStackOfStrings(10);
        while(!s.isFull())
            s.push("first");
        while(!s.isEmpty())
            StdOut.println(s.pop());
    }
}

1.3.2
因为题目有“-”，我就按照FixedCapacityStackOfStrings这个类来解这题了：
was
best
times
of
the
was
the
it

1.3.3
b错，0 1是不可能产生的
f错，1 7 2是不可能产生的
g错，0 2是不可能产生的

1.3.4

import edu.princeton.cs.algs4.In;
import edu.princeton.cs.algs4.Stack;
import edu.princeton.cs.algs4.StdOut;

public class Parentheses{
    private static final char LEFT_PAREN = '(';
    private static final char RIGHT_PAREN = ')';
    private static final char LEFT_BRACE = '{';
    private static final char RIGHT_BRACE = '}';
    private static final char LEFT_BRACKET = '[';
    private static final char RIGHT_BRACKET = ']';

    public static boolean isBalanced(String s){
        Stack<Character> stack = new Stack<Character>();
        for(int i = 0;i < s.length();i++){
            if(s.charAt(i) == LEFT_PAREN) 
                stack.push(LEFT_PAREN);
            if(s.charAt(i) == LEFT_BRACE)
                stack.push(LEFT_BRACE);
            if(s.charAt(i) == LEFT_BRACKET)
                stack.push(LEFT_BRACKET);

            if(s.charAt(i) == RIGHT_PAREN){
                if(stack.isEmpty())
                    return false;
                if(stack.pop() != LEFT_PAREN)
                    return false;
            }else if(s.charAt(i) == RIGHT_BRACE){
                if(stack.isEmpty())
                    return false;
                if(stack.pop() != LEFT_BRACE)
                    return false;
            }else if(s.charAt(i) == RIGHT_BRACKET){
                if(stack.isEmpty())
                    return false;
                if(stack.pop() != LEFT_BRACKET)
                    return false;
            }
        }
        return stack.isEmpty();
    }
    public static void main(String[] args){
        In in = new In();
        String s = in.readAll().trim();
        StdOut.println(isBalanced(s));
    }
}

1.3.5
（书上有答案）
打印N的二进制表示（当N为50 时打印110010）