斐波纳妾数列思维 数组实现
纳妾 哈哈你懂的

//第一种解法:是用数组的形式
public class Solution {
    public int JumpFloor(int target) {
        int[] array = new int[target];
        if(target==1){
            array[target-1] =target;
        }else if(target==2){
            array[target-2] =1;
            array[target-1] = target;
        }
        else{
             array[0] =1;
             array[1] = 2;
            for(int i=3;i<=target;i++){
                array[i-1]=array[i-2]+array[i-3];
            }
        }
        return array[target-1];
    }
}

递归调用实现:

ublic class Solution {
    public int JumpFloor(int target) {
       if(target==1 || target==2){
           return target;
       }
        return JumpFloor(target-1)+JumpFloor(target-2);
    }
}