LeetCode 213: House Robber II

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.

解题思路

如果抢劫第一件房屋，则不可以抢最后一件房屋；否则，可以抢最后一间房屋。因此，可以将环形DP问题转化为两趟线性DP问题，可以复用 House Robber 的代码。另外需要特判一下只有一件房屋的情形。代码如下：

class Solution {private:    int robLinear(vector<int>& nums, int start, int end) {        if (start >= end) return 0;        int pre = nums[start], prePre = 0;        for (int i = start + 2; i <= end; ++i) {            int temp = max(pre, prePre + nums[i-1]);            prePre = pre;            pre = temp;        }        return pre;    }public:    int rob(vector<int>& nums) {        if (nums.size() == 0) return 0;        if (nums.size() == 1) return nums[0];        return max(robLinear(nums, 0, nums.size()-1), robLinear(nums, 1, nums.size()));    }};