Careercup

2014-05-06 01:49

题目链接

原题：

Modify the following code to add a row number for each line is printed



public class Test {
    public static void main(String [] args){
        printParenthesis(3);
    }
    public static void printParenthesis(int n){
        char buffer[] = new char[n*2];
        printP(buffer,0,n,0,0);
    }
    public static void printP(char buffer[], int index, int n, int open, int close){
        if(close == n){
            System.out.println(new String(buffer));
        }else{
            if(open > close){
                buffer[index] = ']';
                printP(buffer, index+1, n, open, close+1);
            }
            if(open < n ){
                buffer[index] = '[';
                printP(buffer,index+1,n,open+1,close);
            }
        }
    }
}

Expected Output:



1.[][][]
2.[][[]]
3.[[]][]
4.[[][]]
5.[[[]]]

What changes needs to be done to accomplish the output expected?

题目：给下面的代码加上一些修改，使得输出的结果能带有序号，如示例中的格式。

解法：代码可能还不太明显，但从结果一看就知道是输出N对括号匹配的所有组合，并且卡塔兰数H(3) = 5，也符合条件。只要加上一个全局的counter，并且在输出语句附近给counter加1，就可以带序号输出了。

代码：

// http://www.careercup.com/question?id=6253551042953216
public class Test {
    static int res_count = 0;
    
    public static void main(String [] args) {
        printParenthesis(3);
    }
    
    public static void printParenthesis(int n) {
        char buffer[] = new char[n * 2];
        res_count = 0;
        printP(buffer, 0, n, 0, 0);
    }
    
    public static void printP(char buffer[], int index, int n, int open, int close) {
        if(close == n) {
            System.out.print((++res_count) + ".");
            System.out.println(new String(buffer));
        } else {
            if (open > close) {
                buffer[index] = ']';
                printP(buffer, index+1, n, open, close + 1);
            }
            if (open < n) {
                buffer[index] = '[';
                printP(buffer, index + 1, n, open + 1, close);
            }
        }
    }
}