一、题目内容

二、题解

解法1：

思路

动态规划
f(0)=1
f(1)=1
f(2)=2
dp[i] = dp[i-1]+dp[i-2]

代码

public class Solution {
    public int jumpFloor(int target) {
        if (target == 0 || target == 1) {
            return 1;
        }
        if (target == 2) {
            return 2;
        }
        return jumpFloor(target - 1) + jumpFloor(target-2);
    }
}

解法2：

思路

双指针
a=1，b=1，c=0
for(int i = 2;i<=target;i++)
c = a + b
a = b
b = c

代码

public class Solution {
    public int jumpFloor(int target) {
        if(target<=1){
            return 1;
        }
        int a = 1,b = 1,c = 0;
        for(int i = 2;i<=target;i++){
            c = a+b;
            a = b;
            b = c;
        }
        return c;
    }
}