假设我将金额存储在 house[k] 中的第 k 个房子.

Suppose I store the amount in kth house in house[k].

假设现在我在 max[k] 中存储了从前 k 个房屋(并且仅前 k 个)中可以掠夺的最大金额.

Suppose now I store the maximum amount of money possible to loot from the first k houses(and first k only) in max[k].

现在考虑没有房子，所以max[0]=0

Now consider no houses, so max[0]=0

现在只考虑第一个房子，max[1]=房子 1 的数量

Now considering only first house, max[1]=amount in house 1

现在考虑前两套房子，

max[2]={either max[1](暗示我们选择掠夺房屋 1)或(房屋 2 的数量 + 我掠夺的最大数量直到房子位于我现在的房子之前 2 个地方)}={max(max[1],house[2]+max[0])}

max[2]={either max[1](implies we chose to loot house 1) or (amount in house 2 + maximum amount that I had looted until the house located 2 places before my current house)}={max(max[1],house[2]+max[0])}

前3个房子类似，max[3]=max(max[2],house[3]+max[1])

观察这个趋势，可以表述为max[k]=max(max[k-1],house[k]+max[k-2]).这个值一直计算到最后当没有更多房子的时候，我们得到前n个房子可以掠夺的最大数量.

Observing this trend, it can be formulated that max[k]=max(max[k-1],house[k]+max[k-2]). This value is calculated till in the end when there are no more houses, we get the maximum amount that can be looted from these first n houses.

DP 问题只有在您之前有过一些练习和熟悉的情况下才会出现，这总是有帮助的.

DP problems strike the head only when you have had some practice and familiarity before, and this always helps.