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Frog Jumping Step One
Problem description : A frog can jump up to 1 level or 2 at a time. Ask the frog to jump on an n-level step
How many jump methods are there in total (different order counts as different results)
如果n=1，只有一种跳法，那就是1
如果n=2，那么有两种跳法，2，[1,1]
如果n=3，那么有三种跳法，[1,1,1],,[1,2],[2,1]
如果n=4，那么有五种跳法，[1,1,1,1],[1,1,2],[1,2,1],[2,1,1],[2,2]
如果n=5，那么有八种跳法，[1,1,1,1,1],[1,1,1,2],[1,1,2,1],[1,2,1,1],[2,1,1,1],[2,2,1],[2,1,2],[1,2,2]
结果为1，2，3，5，8 这不正是斐波那契数列嘛
The problem is converted to solving the Fibonacci sequence. The following are the implementations of recursive and non-recursive functions.
Recursive solution
int jump(int n)
{
if (n == 0 || n == 1 || n == 2)
return n;
else
{
return jump(n - 1) + jump(n - 2);
}
}
Non-recursive solution
int jump(int n)
{
if (n == 0 || n == 1 || n == 2)
{
return n;
}
int a = 1;
int b = 2;
int c = 3;
while (n > 2)
{
c = a + b;
a = b;
b = c;
n--;
}
return c;
}
Frog jumping two steps
Problem description : A frog can jump up to 1 level or 2 levels at a time...it can also jump to n levels. Beg the green
How many jumping methods are there to leapfrog on an n-level step?
问题分析:
f(n) = f(n-1) + f(n-2) + f(n-3) + … + f(n-(n-1)) + f(n-n)= f(0) + f(1) + f(2) + f(3) + … + f(n-2)+f(n-1)
f(n-1) = f(0) + f(1)+f(2)+f(3) + … + f((n-1)-1) = f(0) + f(1) + f(2) + f(3) + … + f(n-2)
so f(n)=2*f(n-1)
Function implementation
int jump(int n)
{
if (n == 0 || n == 1 || n == 2)
{
return n;
}
else
{
return 2 * jump(n - 1);
}
}
Frog jumping three steps
Problem description : A frog can jump up to 1 level or 2 at a time... It can also jump to m level. Beg this
How many ways are there for a frog to jump on an n-level step?
Problem analysis :
f(n) = f(n-1) + f(n-2) + f(n-3) +… + f(nm)
f(n-1) = f(n-2) + f(n-3) + … + f(n-m) + f(n-m-1)
Simplified to get: f(n) = 2f(n-1)-f(nm-1)
Function implementation
int jump(int m,int n)
{
if (n > m)
{
return 2 * jump(m, n - 1) - jump(m, n - m - 1);
}
// n <= m时跟问题二的解法相同
else
{
if (n == 0 || n == 1 || n == 2)
{
return n;
}
else
{
return 2 * jump(m, n - 1);
}
}
}