From 9a16649b10571ee80a7a88cadfbe03bb0a556b55 Mon Sep 17 00:00:00 2001
From: ShusenTang <tangshusen@pku.edu.cn>
Date: Tue, 10 Dec 2019 11:21:59 +0800
Subject: [PATCH] Create 213. House Robber II.md

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+# [213. House Robber II](https://leetcode.com/problems/house-robber-ii/)
+
+# 思路
+这道题是198.House Robber拓展，现在房子排成了一个圆圈，则如果抢了第一家，就不能抢最后一家，因为首尾相连了。
+
+此题的关键点就在于：如果把第一家和最后一家分别去掉，各算一遍能抢的最大值，两个值较大的一个即为所求。
+
+知道了这个关键点后此题就和198题没有任何区别了，198有两个思路，所以此题也有两种思路，时间复杂度均为O(n)，空间复杂度均为O(1)，
+详见[198题解](https://github.com/ShusenTang/LeetCode/blob/master/solutions/198.%20House%20Robber.md)，这里直接给出代码。
+
+# C++
+## 思路一
+``` C++
+class Solution {
+private:
+    int rob_helper(vector<int>& nums, int start, int end){
+        if(start > end) return 0;
+        if(start == end) return nums[end];
+        
+        int tmp, dp1 = nums[start], dp2 = max(nums[start], nums[start + 1]);
+        for(int i = 2; i < end - start + 1; i++){
+            tmp = dp2;
+            dp2 = max(dp2, nums[start + i] + dp1);
+            dp1 = tmp;
+        }
+        return dp2;
+    }
+public:
+    int rob(vector<int>& nums) {
+        int n = nums.size();
+        if (n <= 1) return nums.empty() ? 0 : nums[0];
+        return max(rob_helper(nums, 0, n - 2), rob_helper(nums, 1, n - 1));
+    }
+};
+```
+
+## 思路二
+``` C++
+class Solution {
+private:
+    int rob_helper(vector<int> &nums, int left, int right) {
+        int rob = 0, notRob = 0;
+        for (int i = left; i <= right; i++) {
+            int preRob = rob, preNotRob = notRob;
+            rob = preNotRob + nums[i];
+            notRob = max(preRob, preNotRob);
+        }
+        return max(rob, notRob);
+    }
+public:
+    int rob(vector<int>& nums) {
+        int n = nums.size();
+        if (n <= 1) return nums.empty() ? 0 : nums[0];
+        return max(rob_helper(nums, 0, n - 2), rob_helper(nums, 1, n - 1));
+    }
+};
+```