[Swift]LeetCode746. 使用最小花费爬楼梯 | Min Cost Climbing Stairs

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On a staircase, the i-th step has some non-negative cost cost[i] assigned (0 indexed).

Once you pay the cost, you can either climb one or two steps. You need to find minimum cost to reach the top of the floor, and you can either start from the step with index 0, or the step with index 1.

Example 1:

Input: cost = [10, 15, 20]
Output: 15
Explanation: Cheapest is start on cost[1], pay that cost and go to the top. 

Example 2:

Input: cost = [1, 100, 1, 1, 1, 100, 1, 1, 100, 1]
Output: 6
Explanation: Cheapest is start on cost[0], and only step on 1s, skipping cost[3]. 

Note:

1. cost will have a length in the range [2, 1000].
2. Every cost[i] will be an integer in the range [0, 999].

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数组的每一个索引作为一个阶梯，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，而后你能够选择继续爬一个阶梯或者爬两个阶梯。

您须要找到达到楼层顶部的最低花费。在开始时，你能够选择从索引为 0 或 1 的元素做为初始阶梯。

示例 1:

输入: cost = [10, 15, 20]
输出: 15
解释: 最低花费是从cost[1]开始，而后走两步便可到阶梯顶，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，逐个通过那些1，跳过cost[3]，一共花费6。

注意：

1. cost 的长度将会在 [2, 1000]。
2. 每个 cost[i] 将会是一个Integer类型，范围为 [0, 999]。

---

Runtime: 32 ms

Memory Usage: 18.9 MB

 1 class Solution {
 2     func minCostClimbingStairs(_ cost: [Int]) -> Int {
 3         var f: [Int] = []
 4         f.append(cost[0])
 5         for i in 1...(cost.count + 1) {
 6             var one = f[i - 1]
 7             var two = 0
 8             var now = 0
 9             if i - 2 < 0 {
10                 two = 0
11             } else {
12                 two = f[i - 2]
13             }
14             if i >= cost.count {
15                 now = 0
16             } else {
17                 now = cost[i]
18             }
19             f.append(min(one + now, two + now))
20         }
21         return min(f[cost.count], f[cost.count + 1])
22     }
23 }

---

32ms

 1 class Solution {
 2     func minCostClimbingStairs(_ cost: [Int]) -> Int {
 3         if cost.count == 1 {
 4             return cost[0]
 5         }
 6         if cost.count == 2 {
 7             return min(cost[0], cost[1])
 8         }
 9         
10         var result = Array(repeating: 0, count: cost.count+1)
11         result[0] = 0
12         result[1] = 0
13         for i in 2...cost.count {
14             result[i] = min(result[i-2] + cost[i-2], result[i-1] + cost[i-1])
15         }
16         return result[cost.count]
17     }
18 }

---

36ms

 1 class Solution {
 2     
 3     func minCostClimbingStairs(_ cost: [Int]) -> Int {
 4         var result1 = cost[0]
 5         var result2 = cost[1]
 6         
 7         for i in 2 ..< cost.count {
 8             var base = min(result1, result2)
 9             result1 = result2
10             result2 = cost[i] + base
11         }
12         
13         return min(result1, result2)        
14     }
15 }

---

40ms

 1 class Solution {
 2     func minCostClimbingStairs(_ cost: [Int]) -> Int {
 3         var step1 = 0, step2 = 0
 4         for i in cost.indices.dropFirst(2) {
 5             let step3 = min(step1 + cost[i - 2], step2 + cost[i - 1])
 6             step1 = step2
 7             step2 = step3
 8         }
 9         return min(step1 + cost.dropLast().last!, step2 + cost.last!)
10     }
11 }

---

48ms

 1 class Solution {
 2     func minCostClimbingStairs(_ cost: [Int]) -> Int {
 3         var minCost = cost
 4         for i in 2 ..< cost.count{
 5             minCost[i] += min(minCost[i-1], minCost[i-2])
 6         }
 7         
 8         return min(minCost[cost.count-1], minCost[cost.count-2])
 9     }
10 }

---

64ms

 1 class Solution {
 2     func minCostClimbingStairs(_ cost: [Int]) -> Int {
 3         let n = cost.count
 4         var value1 = cost[0]
 5         var value2 = cost[1]
 6         for i in 2..<n {
 7             (value2, value1) = (min(value2, value1) + cost[i], value2)
 8         }
 9         return min(value1, value2)
10     }
11 }