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第四周作业 #917

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bf095af
Create leetcode_455_024.rb
lglove May 21, 2019
9f5afa1
Create leetcode_746_024.rb
lglove May 21, 2019
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Update NOTE.md
lglove May 21, 2019
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lglove May 21, 2019
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Create leetcode_746_024.rb
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@lglove
lglove authored May 21, 2019
commit 9f5afa1882f272ce9e243df8a886e26869078479
18 changes: 18 additions & 0 deletions Week_04/id_24/leetcode_746_024.rb
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[746: 最小花费爬楼梯](https://leetcode-cn.com/problems/min-cost-climbing-stairs/submissions/)

###解题思路 动态规划问题，主要还是要找到递推公式
### 由题目我们能到看到到达当前位置的最小花费来自于上一步和上上一步的最小花费
### 递归公式就是 f(n) = min(f(n-1) + cost[n-1], f(n-2) + cost(n-2)), f(0) = 0, f[1] = 0,
### f[2] = min(f(1) + cost[1], f(0)+cost[0])

def min_cost_climbing_stairs(cost)
f = []
f[0] = 0
f[1] = 0
f[2] = [f[0] + cost[0], f[1]+cost[1]].min
for i in (2..cost.length) do
f[i] = [f[i-2] + cost[i-2], f[i-1]+cost[i-1]].min
end
return f[cost.length]

end