213. House Robber II

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.

题目含义:198. House Robber的扩展,第一个element 和最后一个element不能同时出现。则分两次call House Robber I. case 1: 不包括最后一个element. case 2: 不包括第一个element.。两者的最大值即为全局最大值

 

复制代码
 1     public int rob(int[] nums) {
 2         if(nums==null || nums.length==0) return 0;  
 3         if(nums.length==1) return nums[0];  
 4         if(nums.length==2) return Math.max(nums[0], nums[1]);  
 5         return Math.max(robsub(nums, 0, nums.length-2), robsub(nums, 1, nums.length-1));         
 6     }
 7     
 8     public int robsub(int[] nums,int start,int end) {
 9         int len = end - start + 1;
10         if (len < 0) return 0;
11         if (len == 1) return nums[start];
12         int[] rt = new int[len];
13         rt[0] = nums[start];
14         rt[1] = Math.max(nums[start], nums[start + 1]);
15         for (int i = 2; i < len; i++) {
16             rt[i] = Math.max(rt[i - 1], rt[i - 2] + nums[start + i]);
17         }
18         return rt[len - 1];
19     }
复制代码

 

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