213. House Robber II

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.

此题和House Robber的不同在于，这个是首尾相连的，需要多考虑一步，当没选中第一个房子的时候可以选择最后一个房子，而当选中第一个房子的时候不可以选择最后一个房子。做法是把数组分成两个部分，然后求这两个部分最大值进行比较，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(nums[0],Math.max(nums[1],nums[2]));

        return Math.max(helper(Arrays.copyOfRange(nums,0,nums.length-1)),helper(Arrays.copyOfRange(nums,1,nums.length)));

    }

    public int helper(int[] nums){

        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-2]+nums[i],dp[i-1]);

        }

        return dp[nums.length-1];

    }

}

这个代码初始值要多判断很多步，下面来一个简练的：

public class Solution {

    public int rob(int[] nums) {

        if(nums.length==0) return 0;

        if(nums.length==1) return nums[0];

        return Math.max(helper(nums,0,nums.length-2),helper(nums,1,nums.length-1));

    }

    public int helper(int[] nums,int start,int end){

        int include = 0;

        int exclude = 0;

        for(int j=start;j<=end;j++){

            int i=include,e=exclude;

            include = e+nums[j];

            exclude = Math.max(i,e);

        }

        return Math.max(include,exclude);

    }

}