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);
}
}
}
}

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