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
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 }
