题目描述:
Note: This is an extension of House Robber.
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.
先写上House Robber.的解答吧,太简单了,就全部一起写了:
public class Solution { public int rob(int[] nums) { if(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]; dp[0]=nums[0]; dp[1]=Math.max(nums[0], nums[1]); for(int i=2;i<nums.length;i++){ dp[i]=Math.max(dp[i-1], dp[i-2]+nums[i]); } return dp[nums.length-1]; } }而本题不允许第一个和最后一个一起被偷,那么我们可以假设小偷从nums[0],偷到nums[n-2],然后又从nums[1]偷到nums[n-1],这样就可以使得nums[0]和nums[n-1]永远不会一起被偷了,然后比较他们的最大值,就是被偷的最大值。
代码如下:
public class Solution { public int rob(int[] nums) { if(nums.length==0) return 0; if(nums.length==1) return nums[0]; if(nums.length==2) return Math.max(nums[0], nums[1]); if(nums.length==3) return Math.max(Math.max(nums[0], nums[1]),nums[2]); int[] dp1=new int[nums.length-1]; dp1[0]=nums[0]; dp1[1]=Math.max(nums[0], nums[1]); for(int i=2;i<nums.length-1;i++){ dp1[i]=Math.max(dp1[i-1], dp1[i-2]+nums[i]); } int[] dp2=new int[nums.length-1]; dp2[0]=nums[1]; dp2[1]=Math.max(nums[1], nums[2]); for(int i=3;i<nums.length;i++){ dp2[i-1]=Math.max(dp2[i-2], dp2[i-3]+nums[i]); } return Math.max(dp1[dp1.length-1], dp2[dp2.length-1]); } }