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1. Tower of Hanoi problem
1. Problem overview
2. Solution analysis
Summarize:
(1) Using disk C as an intermediary, move disks 1 to n-1 from lever A to lever B;
(2) Move the nth disc remaining in the A rod to the C rod;
(3) With lever A as an intermediary; move discs 1 to n-1 from lever B to lever C.
(4) One plate needs to be moved 1 time, two plates need to be moved 3 times, three plates need to be moved 7 times, it is not difficult to come out, n plates need to be moved 2^n -1 times
3. Code implementation
public class TestDemo {
//1个盘子时：A->C 1
//2个盘子时：A->B A->C B->C 3
//3个盘子时：A->C A->B C->B A->C B->A B->C A->C 7
public static void move(char pos1,char pos2){
//模拟鼠标操作
System.out.print(pos1+"->"+pos2+" ");
}
/*
n 代表盘子个数
pos1 盘子的起始位置
pos2 盘子的中转位置
pos3 盘子的结束位置
*/
public static void hanoi(int n,char pos1,char pos2,char pos3){
if(n==1){
move(pos1,pos3);
}else{
hanoi(n-1,pos1,pos3,pos2);//n-1次的结束位置应该是B上面，除了最下面的盘子，其他全被移到了B上
move(pos1,pos3);//此时A上面仅剩下最下面的盘子，只需要将它移到C上
hanoi(n-1,pos2,pos1,pos3);//此时盘子数是没有最下面的盘子，我们仅需要把B上的盘子通过A移动到C上
}
}
public static void main(String[] args) {
hanoi(1,'A','B','C');
System.out.println();
hanoi(2,'A','B','C');
System.out.println();
hanoi(3,'A','B','C');
System.out.println();
hanoi(4,'A','B','C');
System.out.println();
hanoi(5,'A','B','C');
}
}
2. Frog jumping steps
1. Problem overview
A frog can jump up 1 steps at a time, or 2 steps at a time. Find how many ways the frog can jump up a n steps
2. Solution analysis
3. Recursive implementation
public class TestDemo2 {
public static int frogJump(int n){//台阶数量
if(n==1 || n==2){
return n;
}else{
return frogJump(n-1)+frogJump(n-2);
}
}
public static void main(String[] args) {
System.out.println(frogJump(1));
System.out.println(frogJump(2));
System.out.println(frogJump(3));
System.out.println(frogJump(4));
System.out.println(frogJump(5));
}
}
4. Iterative implementation
public class TestDemo2 {
public static int frogJump(int n){//台阶数量
if(n==1 || n==2){
return n;
}
int f1 = 1;
int f2 = 2;
int f3 = 0;
for (int i = 3; i <=n ; i++) {
f3 = f1+f2;
f1 = f2;
f2 = f3;
}
return f3;
}
public static void main(String[] args) {
System.out.println(frogJump(1));
System.out.println(frogJump(2));
System.out.println(frogJump(3));
System.out.println(frogJump(4));
System.out.println(frogJump(5));
}
}
