【一天一大 lee】使用最小花费爬楼梯 (难度:简单) - Day20201221

输入: cost = [10, 15, 20]
输出: 15
解释: 最低花费是从cost[1]开始，然后走两步即可到阶梯顶，一共花费15。

输入: cost = [1, 100, 1, 1, 1, 100, 1, 1, 100, 1]
输出: 6
解释: 最低花费方式是从cost[0]开始，逐个经过那些1，跳过cost[3]，一共花费6。

/**
 * @param {number[]} cost
 * @return {number}
 */
var minCostClimbingStairs = function(cost) {
    const n = cost.length
    const dp = new Array(n + 1)
    dp[0] = dp[1] = 0
    for (let i = 2; i <= n; i++) {
        dp[i] = Math.min(dp[i - 1] + cost[i - 1], dp[i - 2] + cost[i - 2])
    }
    return dp[n]
}

var minCostClimbingStairs = function(cost) {
    const n = cost.length
    let prevNum = 0,
        indexNum = 0
    for (let i = 2; i <= n; i++) {
        let next = Math.min(indexNum + cost[i - 1], prevNum + cost[i - 2])
        prevNum = indexNum
        indexNum = next
    }
    return indexNum
}