LeetCode 213. House Robber II

213.
House Robber II

My
Submissions QuestionEditorial Solution

Total
Accepted: 27976 Total Submissions: 91078 Difficulty: Medium

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.

这题是上一题House
Robber的升级版~~新加了环形的街道biu biu biu~~

所以就会考虑到，最后一个和第一个房子是不能够同时进入的~要不然会告诉警察叔叔~~

所以分为两种情况~

0.不包括最后一个屋子~就抢劫0~n-2号屋子~

1.不包括第一个屋子~就抢劫1~n-1号屋子~

这样的话，return上面两种情况的最大值就好了~调用两次子函数求值，主函数取其最大值返回~

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]);
return max(func(nums, 0, n-2), func(nums, 1, n-1));
}
int func(vector<int>& nums, int begin, int end) {
int n = end - begin + 1;
vector<int> dp(n);
dp[0] = nums[begin];
dp[1] = max(nums[begin], nums[begin+1]);
for(int i = 2; i < n; i++) {
int temp = dp[i - 2] + nums[begin+i];
dp[i] = max(temp, dp[i-1]);
}
return dp[n - 1];
}
};