house robber II

fishercoder1534 · fishercoder1534 · commit 89a68de661ed · 2016-08-01T21:36:39.000-07:00
diff --git a/MEDIUM/src/medium/HouseRobberII.java b/MEDIUM/src/medium/HouseRobberII.java
@@ -0,0 +1,45 @@
+package medium;
+/**213. House Robber II  QuestionEditorial Solution  My Submissions
+Total Accepted: 34216
+Total Submissions: 107734
+Difficulty: Medium
+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.*/
+public class HouseRobberII {
+
+    /**Another dp problem:
+     * separate them into two cases:
+     * 1. rob from house 1 to n-1, get max1
+     * 2. rob from house 2 to n, get max2
+     * take the max from the above two max*/
+    public int rob(int[] nums) {
+        if(nums == null || nums.length == 0) return 0;
+        if(nums.length == 1) return nums[0];
+        if(nums.length == 2) return Math.max(nums[0], nums[1]);
+        int[] dp = new int[nums.length-1];
+        
+        //rob 1 to n-1
+        dp[0] = nums[0];
+        dp[1] = Math.max(nums[0], nums[1]);
+        for(int i = 2; i < nums.length-1; i++){
+            dp[i] = Math.max(dp[i-2] + nums[i], dp[i-1]);
+        }
+        int prevMax = dp[nums.length-2];
+        
+        //rob 2 to n
+        dp = new int[nums.length-1];
+        dp[0] = nums[1];
+        dp[1] = Math.max(nums[1], nums[2]);
+        for(int i = 3; i < nums.length; i++){
+            dp[i-1] = Math.max(dp[i-3] + nums[i], dp[i-2]);
+        }
+        return Math.max(prevMax, dp[nums.length-2]);
+    }
+
+}
