我刚刚计算完 OBST 的平均成本,我知道我计算正确。我的下一个任务是按顺序打印树。我尝试使用递归,但似乎无法摆脱空指针错误。
这是我的代码:
public class OBST {
static String[] keysA;
static Integer[][] root;
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int tot = sc.nextInt();
HashMap<String, Double> hm = new HashMap<String, Double>();
int uniqNum = 0;
String[] rawInput = new String[tot];
for(int i=0; i<tot; i++) {
String tmp1 = sc.next();
if(i==0) {
hm.put(tmp1, 1.0);
uniqNum += 1.0;
} else if( i != 0) {
if(!hm.containsKey(tmp1)) {
hm.put(tmp1, 1.0);
uniqNum += 1.0;
} else {
Double tmpfreq = 0.0;
tmpfreq = hm.get(tmp1);
hm.put(tmp1, (tmpfreq + 1.0));
}
}
}
Set<String> keys = hm.keySet();
keysA = keys.toArray(new String[uniqNum]);
Double[] freqsA = new Double[uniqNum];
Arrays.sort(keysA);
for(int i=0; i<uniqNum; i++) {
Double tmp = 0.0;
String tmpK = keysA[i];
tmp = hm.get(tmpK);
tmp = tmp/tot;
freqsA[i] = tmp;
}
Double[][] eee = new Double[uniqNum+2][uniqNum+1];
Double[][] www = new Double[uniqNum+2][uniqNum+1];
//matrix to store optimal structure
root = new Integer[uniqNum+1][uniqNum+1];
for(int i=1; i<uniqNum+2; i++) {
eee[i][i-1] = 0.0;
www[i][i-1] = 0.0;
}
for(int l=1; l<uniqNum+1; l++) {
for(int i=1; i<=uniqNum-l+1; i++) {
int j = i + l - 1;
eee[i][j] = Double.MAX_VALUE;
www[i][j] = www[i][j-1] + freqsA[j-1];
for(int r=i; r<=j; r++) {
Double t = eee[i][r-1] + eee[r+1][j] + www[i][j];
if(t<eee[i][j]) {
eee[i][j] = t;
root[i][j] = r-1;
}
}
}
}
//total cost
System.out.println(eee[1][uniqNum]);
printTree(1,uniqNum-1,-1, "");
}
public static void printTree(int min, int max, int parent, String s) {
int r = root[min][max];
if(parent == -1 ) {
System.out.println(keysA[r] + " is root");
} else if(min < parent) {
System.out.println(keysA[r] + " is the left child of " + s);
} else {
System.out.println(keysA[r] + " is the right child of " + s);
} if(min < max) {
printTree(min,r,r+1,keysA[r]);
printTree(r+1,max,r,keysA[r]);
}
}
}
我的麻烦在于方法打印树。