题解 | #最小代价爬楼梯#Erlang状态转移关系

解题思路

通过第n个阶梯的花费为通过前两个阶梯中累计花费最小的那个并加上自身的花费，即cost(n) = min(cost(n-1), cost(n-2)) + cost

最终计算得出cost(last) = min(cost(last - 1), cost(last -2))

代码

-spec min_cost_climbing_stairs(Cost :: [integer()]) -> integer().
min_cost_climbing_stairs(Cost) ->
    do_min_cost_climbing_stairs(Cost, _TotalCost = #{}, _Args = #{index => 1}).

do_min_cost_climbing_stairs([Cost | T], TotalCost, Args = #{index := Index}) ->
    PreCost = maps:get(Index - 1, TotalCost, 0),
    PrePreCost = maps:get(Index - 2, TotalCost, 0),
    NewCost = min(PreCost, PrePreCost) + Cost,
    do_min_cost_climbing_stairs(T, TotalCost#{Index => NewCost}, Args#{index := Index + 1});
do_min_cost_climbing_stairs(_, TotalCost, _Args = #{index := Index}) ->
    PreCost = maps:get(Index - 1, TotalCost, 0),
    PrePreCost = maps:get(Index - 2, TotalCost, 0),
    min(PreCost, PrePreCost).