refactor 213

fishercoder1534 · fishercoder1534 · commit 1a41e759c62b · 2018-09-22T08:11:42.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_213.java b/src/main/java/com/fishercoder/solutions/_213.java
@@ -1,52 +1,55 @@
 package com.fishercoder.solutions;
-/**213. House Robber II
+/**
+ * 213. House Robber II
 
  Note: This is an extension of House Robber.
 
-After robbing those houses on that street,
+ 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.
+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.
 */
 public class _213 {
+    public static class Solution1 {
+        /**
+         * 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];
 
-    /**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 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]);
+            //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]);
         }
-        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]);
     }
-
 }
