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 }
分类:
leetcode_dp
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