chore: add daily leetcode post [213]打家劫舍 II_translated

Author: longsizhuo

app/docs/CommunityShare/Leetcode/[213]打家劫舍 II_translated.md

@@ -0,0 +1,42 @@
+---
+title: 213.Hiccup II
+date: '2024.01.01 0:00'
+tags:
+  - Python
+  - answer
+  - Array
+  - Dynamic planning
+abbrlink: 85beb0bf
+---
+# topic：
+
+
+[topic链接](https://leetcode-cn.com/problems/house-robber-ii/)
+
+# Thought：
+这一次终于懂一点Dynamic planning了，The first is the youngest problem，I first thought it was to consider where to start，Start from the first or the second one（这刚好是和Hiccup1Difference）。
+但实际上最小子问题还是和Hiccup1Same：Whether to rob the current house。
+First we set one`dp`Array，`dp[i]`Indicate robbery to the first`i`The maximum return when a house。
+Then we rob the current house after we rob the current house `dp[i] = dp[i-2] + nums[i]`，(Because the house can not be robbed before the current house is robbed)，Do not rob the current house `dp[i] = dp[i-1]`。
+So we can get the state transfer equation：`dp[i] = max(dp[i-2] + nums[i], dp[i-1])`。
+Last classification discussion and robbery, the first or the second start。
+
+# Code：
+
+```python
+class Solution:
+    def rob(self, nums: List[int]) -> int:
+        def rob1(nums: List[int]) -> int:
+            if len(nums) == 1:
+                return nums[0]
+            ans = 0
+            dp = [0] * len(nums)
+            # def: dp[i]Express the robberyiHome
+            # theft FirstiHouse，The income isdp[i-2]+nums[i], 不theftFirstiHouse，The income isdp[i-1]
+            for i in range(len(nums)):
+                dp[i] = max(dp[i - 2] + nums[i], dp[i - 1])
+                ans = max(ans, dp[i])
+            return ans
+        # 打劫First一家，或者First二家开始
+        return max(rob1(nums[:-1]), rob1(nums[1:])) if len(nums) != 1 else nums[0]
+```