我的任務是添加一堆打印語句,以顯示Tower of Hanoi的完整輸出,以查看並理解它在幕後做了什麼,而不僅僅是給出最終結果。河內塔顯示輸出。如何顯示第一次遞歸調用的1個選項卡,第二次遞歸調用的2個選項卡等?
class TowersApp {
static int nDisks = 3;
public static void main(String[] args) {
doTowers(nDisks, 'A', 'B', 'C');
}
public static void doTowers(int topN, char from, char inter, char to) {
int i = 0;
if(topN==1) {
System.out.println("Enter (" + topN + " disk): " + "s=" + from + ", i=" + inter + ", d=" + to);
System.out.println("Base case: move disk " + topN + " from " + from + " to "+ to);
System.out.println("Return (" + topN + " disk)"); }
else {
System.out.println("Enter (" + topN + " disks): " + "s=" + from + ", i=" + inter + ", d=" + to);
doTowers(topN-1, from, to, inter);
System.out.println("Move bottom disk " + topN +
" from " + from + " to "+ to);
doTowers(topN-1, inter, from, to);
System.out.println("Return (" + topN + " disks)");
}
}
}
這是我的輸出。我唯一缺少的是縮進。我需要有1片遞歸的第一級,2個標籤遞歸的第二級等等......這裏就是我的意思是:
電流輸出:
Enter (3 disks): s=A, i=B, d=C
Enter (2 disks): s=A, i=C, d=B
Enter (1 disk): s=A, i=B, d=C
Base case: move disk 1 from A to C
Return (1 disk)
Move bottom disk 2 from A to B
Enter (1 disk): s=C, i=A, d=B
Base case: move disk 1 from C to B
Return (1 disk)
Return (2 disks)
Move bottom disk 3 from A to C
Enter (2 disks): s=B, i=A, d=C
Enter (1 disk): s=B, i=C, d=A
Base case: move disk 1 from B to A
Return (1 disk)
Move bottom disk 2 from B to C
Enter (1 disk): s=A, i=B, d=C
Base case: move disk 1 from A to C
Return (1 disk)
Return (2 disks)
Return (3 disks)
所需的輸出:
Enter (3 disks): s=A, i=B, d=C
Enter (2 disks): s=A, i=C, d=B
Enter (1 disk): s=A, i=B, d=C
Base case: move disk 1 from A to C
Return (1 disk)
Move bottom disk 2 from A to B
Enter (1 disk): s=C, i=A, d=B
...................................
我需要某種計數器來「計數」我進入函數的次數嗎?但是,那甚至可以用遞歸?也許我在分析什麼時候有更簡單的解決方案來解決這個問題?
這是一個非常愚蠢的問題,但我如何打印\ t深度= 1,\ t \ t深度= 2,\ t \ t \ t深度= 3? – 2010-11-22 16:29:01