New Problem "House Robber II"

README.md

@@ -7,6 +7,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|197|[House Robber II](https://leetcode.com/problems/house-robber-ii/)| [C++](./algorithms/houseRobber/houseRobber.II.cpp)|Medium|
 |196|[Word Search II](https://leetcode.com/problems/word-search-ii/)| [C++](./algorithms/wordSearch/wordSearch.II.cpp)|Hard|
 |195|[Add and Search Word - Data structure design](https://leetcode.com/problems/add-and-search-word-data-structure-design/)| [C++](./algorithms/addAndSearchWord/AddAndSearchWord.cpp)|Medium|
 |194|[Course Schedule II](https://leetcode.com/problems/course-schedule-ii/)| [C++](./algorithms/courseSchedule/CourseSchedule.II.cpp)|Medium|

algorithms/houseRobber/houseRobber.II.cpp

@@ -0,0 +1,57 @@
+// Source : https://leetcode.com/problems/house-robber-ii/
+// Author : Hao Chen
+// Date   : 2015-06-10
+
+/********************************************************************************** 
+ * 
+ * Note: This is an extension of House Robber.
+ * 
+ * After robbing those houses on that street, the thief has found himself a new place 
+ * for his thievery so that he will not get too much attention. This time, all houses 
+ * at this place are arranged in a circle. That means the first house is the neighbor 
+ * of the last one. Meanwhile, the security system for these houses remain the same as
+ * for those in the previous street. 
+ * 
+ * Given a list of non-negative integers representing the amount of money of each house, 
+ * determine the maximum amount of money you can rob tonight without alerting the police.
+ * 
+ * Credits:Special thanks to @Freezen for adding this problem and creating all test cases.
+ *               
+ **********************************************************************************/
+
+
+class Solution {
+public:
+    int orginal_rob(vector<int> &money, int start, int end) {
+        int n2=0; 
+        int n1=0; 
+        
+        for (int i=start; i<end; i++){
+            int current = max(n1, n2 + money[i]);
+            n2 = n1;
+            n1 = current;
+        }
+        return n1;
+    }
+    
+    int rob(vector<int>& nums) {
+        int n = nums.size();
+        switch (n) {
+            case 0:
+                return 0;
+            case 1:
+                return nums[0];
+            case 2:
+                return max(nums[0], nums[1]);
+            default:
+                /*
+                 * the idea is we cannot rob[0] and rob[n-1] at same time
+                 * so, we rob [0 .. n-2] or [1 .. n-1], can return the maxinum one.
+                 */
+                int m1 = orginal_rob(nums, 0, n-1);
+                int m2 = orginal_rob(nums, 1, n);
+                
+                return max(m1, m2);
+        }
+    }
+};