题目：
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.
思路：
This is an extension of House Robber. There are two cases here 1) 1st element is included and last is not included 2) 1st is not included and last is included. Therefore, we can use the similar dynamic programming approach to scan the array twice and get the larger value.
代码：
1 public int rob(int[] nums) { 2 if(nums==null||nums.length==0) 3 return 0; 4 5 int n = nums.length; 6 7 if(n==1){ 8 return nums[0]; 9 } 10 if(n==2){ 11 return Math.max(nums[1], nums[0]); 12 } 13 14 //include 1st element, and not last element 15 int[] dp = new int[n+1]; 16 dp[0]=0; 17 dp[1]=nums[0]; 18 19 for(int i=2; i<n; i++){ 20 dp[i] = Math.max(dp[i-1], dp[i-2]+nums[i-1]); 21 } 22 23 //not include frist element, and include last element 24 int[] dr = new int[n+1]; 25 dr[0]=0; 26 dr[1]=nums[1]; 27 28 for(int i=2; i<n; i++){ 29 dr[i] = Math.max(dr[i-1], dr[i-2]+nums[i]); 30 } 31 32 return Math.max(dp[n-1], dr[n-1]); 33 }
运行结果：
输入nums={1,4,5,2,9,8}
1. 包括1st，不包括last one
dp[0]=0
dp[1]=nums[0]=1
dp[2]=4, (dp[1], dp[0]+nums[1])
dp[3]=6, (dp[2], dp[1]+nums[2])
dp[4]=6, (dp[3], dp[2]+nums[3])
dp[5]=15, (dp[4], dp[3]+nums[4])
2. 不包括1st，包括last one
dr[0]=0
dr[1]=4
dr[2]=5, (dr[1], dr[0]+nums[2])
dr[3]=6, (dr[2], dr[1]+nums[3])
dr[4]=14, (dr[3], dr[2]+nums[4])
dr[5]=14, (dr[4], dr[3]+nums[5])
Reference: http://www.programcreek.com/2014/05/leetcode-house-robber-ii-java/
解法二：
public class Solution { public int rob(int[] nums) { if(nums==null||nums.length==0) return 0; int n = nums.length; if(n==1){ return nums[0]; } if(n==2){ return Math.max(nums[1], nums[0]); } //include 1st element, and not last element int[] dp = new int[n+1]; dp[0]=nums[0]; dp[1]=nums[0]; //include 1st element, so doesnt consider the 2nd element for(int i=2; i<n; i++){ dp[i] = Math.max(dp[i-1], dp[i-2]+nums[i]); } //not include frist element, and include last element int[] dr = new int[n+1]; dr[0]=0; dr[1]=nums[1]; for(int i=2; i<n; i++){ dr[i] = Math.max(dr[i-1], dr[i-2]+nums[i]); } return Math.max(dp[n-2], dr[n-1]); //dp[n-1] include the last element, dp[n-2] is correct. } }
转载于:https://www.cnblogs.com/hygeia/p/4647856.html
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