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); } }