114 - 746使用最小花费爬楼梯

题目

数组的每个索引做为一个阶梯，第 i个阶梯对应着一个非负数的体力花费值 cost[i](索引从0开始)。

每当你爬上一个阶梯你都要花费对应的体力花费值，然后你可以选择继续爬一个阶梯或者爬两个阶梯。

您需要找到达到楼层顶部的最低花费。在开始时，你可以选择从索引为 0 或 1 的元素作为初始阶梯。

示例 1:

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。

示例 2:

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。

注意：

1. cost 的长度将会在 [2, 1000]。

2. 每一个 cost[i] 将会是一个Integer类型，范围为 [0, 999]。

解答

https://leetcode-cn.com/problems/min-cost-climbing-stairs/solution/python3dong-tai-gui-hua-by-pandawakaka-3/

/**
 * @param {number[]} cost
 * @return {number}
 */
var minCostClimbingStairs = function(cost) {
  let a = cost[0],
    b = cost[1]
  for (let i = 2; i < cost.length; i++) {
    let next = Math.min(a, b) + cost[i];
    a = b
    b = next
  }
  return Math.min(a, b)
};

Runtime: 64 ms, faster than 44.16% of JavaScript online submissions for Min Cost Climbing Stairs.

Memory Usage: 35 MB, less than 87.50% of JavaScript online submissions for Min Cost Climbing Stairs.

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        a = cost[0]
        b = cost[1]
        for i in range(2, len(cost)):
            _next = min(a, b)+cost[i]
            a = b
            b = _next
        return min(a, b)

Runtime: 60 ms, faster than 97.16% of Python3 online submissions for Min Cost Climbing Stairs.

Memory Usage: 14 MB, less than 7.69% of Python3 online submissions for Min Cost Climbing Stairs.

func min(a, b int) int {
    if a < b {
        return a
    }
    return b
}
func minCostClimbingStairs(cost []int) int {
    a := cost[0]
    b := cost[1]
    for i := 2; i < len(cost); i++ {
        next := min(a, b) + cost[i]
        a = b
        b = next
    }
    return min(a, b)
}

Runtime: 4 ms, faster than 86.86% of Go online submissions for Min Cost Climbing Stairs.

Memory Usage: 2.9 MB, less than 100.00% of Go online submissions for Min Cost Climbing Stairs.

看到了一个python的三行解法

https://leetcode.com/problems/min-cost-climbing-stairs/discuss/144682/3-Lines-Java-Solution-O(1)-space

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        for i in range(2, len(cost)):
            cost[i] += min(cost[i-1], cost[i-2])
        return min(cost[-1], cost[-2])

Runtime: 68 ms, faster than 64.36% of Python3 online submissions for Min Cost Climbing Stairs.

Memory Usage: 13.9 MB, less than 7.69% of Python3 online submissions for Min Cost Climbing Stairs.

果然效率上要慢了很多。。