题解 | #最小花费爬楼梯#

let buf = ''
process.stdin.on('readable', () => {
    let chunk = process.stdin.read()
    if (chunk) buf += chunk.toString()
})
let n
let cost = []
let dp = []
process.stdin.on('end', () => {
    buf.split('\n').forEach((line, i) => {
        if (i === 0) n = Number(line)
        else if (i === 1) cost = line.split(' ').map(Number)
        else {

            dp[0] = 0
            dp[1] = 0

            for (let s = 2; s <= n; s++) {
                dp[s] = Math.min(dp[s - 1] + +cost[s - 1], dp[s - 2] + cost[s - 2])
            }

            console.log(dp[n])
        }
    })
})

解析

dp[x]代表第x台阶的已花费，状态转移方程：dp[x] = Math.min(dp[x - 1] + cost[x - 1], dp[x - 2] + cost[x - 2])

说明：dp[x - 1]一步之前的花费，cost[x - 1]如果真的走了这个台阶的花费

第一次自己做出dp好激动