<h3><a title="题目描述" href="https://leetcode-cn.com/problems/house-robber/">题目描述</a></h3>  <h4>解法：</h4> <pre class="EnlighterJSRAW" data-enlighter-language="null">class Solution: def rob(self, nums: List[int]) -> int: cur, pre = 0,0 for num in nums: cur,pre = max(cur,pre + num),cur return cur</pre>   #### Note - 动态规划好像比之前会写一些了 但还是菜。 - [简单易懂的题解分析](https://leetcode-cn.com/problems/house-robber/solution/da-jia-jie-she-dong-tai-gui-hua-jie-gou-hua-si-lu-/ "简单易懂的题解分析") - **重点在思考状态以及转移方程如何定义** - 又懒又菜。 - 有一个很有意思的点不得不提 ~~当然也有我太菜的问题~~ <pre class="EnlighterJSRAW" data-enlighter-language="null"> cur = max(cur,pre + num) pre = cur </pre>   其实我原先的思路转换成代码时是这么写的，果然错了-。- <pre class="EnlighterJSRAW" data-enlighter-language="null"> cur = 5 pre = 0 num = 6 cur,pre = max(cur, pre+num) ,cur //输出cur = 6,pre = 5 //错误的情况： cur = max(cur,pre+num) //cur = 6 pre = cur //pre=6 </pre> 也就是说要做到同时赋值还是用a,b = c,d 比较合理