213. House Robber II

Problem:

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.

Analysis:

把first 和last element分别去掉求house robber。然后取2个house robber的最大值。实现houseRobber花了点时间，因为start和end是动态的。记住一点就好，在循环里面F[i%2]的第一个肯定是i%2==0。这样一开始的i就必须是2，不能是动态的。

Solution:
第二遍实现感觉编码能力提升太多。

class Solution {
    public int rob(int[] nums) {
        if (nums == null || nums.length == 0)
            return 0;
        if (nums.length() == 1) return nums[0];
        return Math.max(helper(nums, 2, nums.length), helper(nums, 1, nums.length - 1));
    }

    private int helper(int[] nums, int start, int end) {
        int[] dp = new int[nums.length + 1];
        dp[start - 1] = 0;
        dp[start] = nums[start - 1];
        for (int i = start + 1; i <= end; i++) {
            dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i - 1]);
        }
        return dp[end];
    }
}