题解 | #跳台阶扩展问题#

这个就是最后多了个它自己，最后++就可

// count(1) = 1
// count(2) = 2
// count(3) = 3
// count(n) = count(n - 1) + count(n - 2) + ... + count(1) + 1
function count(n){
    if(n < 3){
        return n
    }
    let dp = [0,1,2]
    for(let i = 3;i <= n;i++){
        dp[i] = 0
        for(let j = i - 1;j > 0;j--){
            dp[i] += dp[j]
        }
        dp[i]++
    }
    return dp[n]
}

console.info(count(n))