leetcode 213. House Robber II 入室抢劫 抢劫问题 + 一道经典的DP动态规划问题

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.

本题是上一道题leetcode 740. Delete and Earn 动态规划DP、leetcode 198. House Robber入室抢劫 DP求解 和 leetcode 337. House Robber III DP动态规划 + DFS深度有限遍历 的一个简单的变形，不过上一题是一个数组，本题是一个环。

这个问题转化为上一道题的问题就是，因为第一个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.

代码如下：

/*
 * http://blog.csdn.net/xudli/article/details/45886721
 * 
 * */
public class Solution 
{
    public int rob(int[] nums) 
    {
        //特殊情况处理
        if(nums==null || nums.length<=0)
            return 0;
        else if(nums.length==1)
            return nums[0];
        else if(nums.length==2)
            return Math.max(nums[0], nums[1]);
        else 
        {
            /*
             * 因为第一个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
             * 
             * */
            int a=robChoose(nums,0,nums.length-2);
            int b=robChoose(nums,1,nums.length-1);
            return Math.max(a, b);
        }
    }

    private int robChoose(int[] nums, int begin, int end) 
    {
        int len=end-begin+1;
        int []dp=new int[len];
        dp[0]=nums[begin];
        dp[1]=Math.max(nums[begin], nums[begin+1]);
        for(int i=2;i1], nums[begin+i]+dp[i-2]);
        }
        return dp[len-1];
    }
}

下面是C++的做法，就是做两次DP动态规划即可

代码如下：

#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

class Solution 
{
public:
    int rob(vector<int>& nums)
    {
        if (nums.size() <= 0)
            return 0;
        else if (nums.size() == 1)
            return nums[0];

        int a = getRes(nums, 0, nums.size() - 2);
        int b = getRes(nums, 1, nums.size() - 1);
        return max(a, b);
    }

    int getRes(vector<int>a, int beg, int end)
    {
        int len = end - beg + 1;
        vector<int>dp(len, 0);
        if (len == 0)
            return 0; 
        dp[0]=a[beg];
        dp[1] = max(a[beg], a[beg + 1]);
        for (int i = beg + 2; i <= end; i++)
            dp[i-beg] = max(dp[i-beg - 1], a[i] + dp[i-beg - 2]);
        return dp[len - 1];
    }
};