Leetcode 213. House Robber II

问题描述

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.

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

问题分析：这个问题是House Robber I的变种或者升级，将房子的布局由原来的一排变换成现在的环形。环形就意味着多了一个约束：最后一个房子和第一个房子不能同时选择。可以分情况进行讨论，1、不选择第一个房子，2、不选择最后一个房子。所以可以调用两次House RobberI的算法求解。注意特殊对待n=1的情况。
代码如下：

    public int rob(int[] nums) {
        if(nums==null)
            return 0;
        int n=nums.length;
        if(n==0)
            return 0;
        if(n==1)
            return nums[0];
        return Math.max(myrob(nums,0,n-1),myrob(nums,1,n-1));
    }
    public int myrob(int[] nums,int st,int len){
        int n=len;
        int[]dprob=new int[n];
        int[]dpnrob=new int[n];
        dprob[0]=nums[st];
        dpnrob[0]=0;
        for(int i=1;i<n;i++){
            dprob[i]=dpnrob[i-1]+nums[st+i];
            dpnrob[i]=Math.max(dprob[i-1],dpnrob[i-1]);
        }
        return Math.max(dprob[n-1],dpnrob[n-1]);
    }