解题思路:见跳台阶扩展问题 讨论榜第一
function jumpFloorII(number)
{
// write code here
if(number === 1 || number === 0){
return 1;
}else{
return 2 * jumpFloorII(number - 1);
}
}
解题思路:见跳台阶扩展问题 讨论榜第一
function jumpFloorII(number)
{
// write code here
if(number === 1 || number === 0){
return 1;
}else{
return 2 * jumpFloorII(number - 1);
}
}