動態規划之青蛙跳台階
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//一只青蛙，一次只能跳一級台階或兩級台階，問這個青蛙跳上n級台階有多少種跳法？
//核心思想：fn(n) = fn(n - 1) + fn(n - 2)
function jump(n){
if(n <= 0){
return 0;
}else if(n == 1){
return 1;
}else if(n == 2){
return 2;
}
return jump(n - 1) + jump(n - 2);
}
console.log(jump(4));
//一只青蛙，一次只能跳一級台階或兩級台階或n級台階，問這個青蛙跳上n級台階有多少種跳法？
//核心思想：fn(n) = fn(n - 1) + fn(n - 2) + ... + fn(2) + fn(1)
function jump2(n){
if(n <= 0){
return 0;
}else if(n == 1){
return 1;
}else if(n == 2){
return 2;
}
var result = 0;
for(var i = 1; i < n; i++){
result += jump2(n - i);
}
return result + 1;//+1表示從0級台階直接跳上n級台階
}
console.log(jump2(3));
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