Leetcode 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.

本题在 House Robber 的基础上加上了首尾两间房子相连这一条件。为了运用 House Robber 的解题方法，将此题分解为两种情况：[0,n-2] 和 [1,n-1] , 在这两种情况中选较大者，这两种情况每种的解法和 House Robber 方法一致（DP）。

class Solution {
public:
int rob(vector<int>& nums) {
int n = nums.size();
if (n < 2) return n ? nums[0] : 0;
return max(robber(nums, 0, n - 2), robber(nums, 1, n - 1));
}
private:
int robber(vector<int>& nums, int l, int r) {
int pre = 0, cur = 0;
for (int i = l; i <= r; i++) {
int temp = max(pre + nums[i], cur);
pre = cur;
cur = temp;
}
return cur;
}
};