LeetCode-House Robber II-解题报告

原题链接：https://leetcode.com/problems/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.

和house robber不同的地方在于首位也算相邻，只需要将环状dp变成线性dp

1 - n dp一次。

0 - n-1 dp一次。

取两次两次dp的最大值 

我开始没想清楚，用了3次dp，不清不楚就ac了。。。

class Solution {
public:
int rob(vector<int>& nums) {
int n = nums.size();
if (n == 0)return 0;
if (n == 1)return nums[0];
if (n == 2)return max(nums[0], nums[1]);
if (n == 3)return max(max(nums[0], nums[1]), nums[2]);
int prepre = nums[0], t;
int pre = nums[1];
pre = max(prepre, pre);
int ans;
for (int i = 2; i < n - 1; ++i)
{
t = pre;
pre = max(prepre + nums[i], pre);
prepre = t;
}
ans = pre;
prepre = nums[1];
pre = nums[2];
pre = max(prepre, pre);
for (int i = 3; i < n; ++i)
{
t = pre;
pre = max(prepre + nums[i], pre);
prepre = t;
}
ans = max(pre, ans);

prepre = nums[2];
pre = nums[3];
pre = max(prepre, pre);
for (int i = 4; i < n; ++i)
{
t = pre;
pre = max(prepre + nums[i], pre);
prepre = t;

}
ans = max(pre, ans);
return ans;
}
};