Leetcode 213. House Robber II(python+cpp)

题目

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.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.

Example 1:
Input: [2,3,2]
Output: 3
Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2),because they are adjacent houses.

Example 2:
Input: [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).Total amount you can rob = 1 + 3 = 4.

解法：动态规划

解析：这道题目其实难度不大，对198的解题思路稍作改进就可以。与198的唯一不同之处在于，这边成环，也就是说第一幢房子和最后一幢房子相邻。那么只需要分类讨论两种情况即可，也就是分别包括第一幢房子和最后一幢房子进行动规的最大值。
具体做法为，排除第一幢房子，求nums[1:]的动态规划最大值，和排除最后一幢房子，求nums[:-1]的动态规划值即可。具体动态规划的过程与Leetcode 198完全一样，请参考Leetcode 198 解法.唯一需要注意的是对数组长度为1的情况下的分开讨论

class Solution:def rob(self, nums: List[int]) -> int:def rob_sub(new_nums):n = len(new_nums)if n==0: return 0dp = [0]*(n+1)dp[1] = new_nums[0]for i in range(2,n+1):dp[i] = max(dp[i-1],dp[i-2]+new_nums[i-1])return dp[-1]if len(nums) == 1:return nums[0]return max(rob_sub(nums[:-1]),rob_sub(nums[1:]))

时间复杂度：O(2N)=O(N)
空间复杂度：O(2N)=O(N)

C++版本：特别注意C++中的vector是如何切片的

class Solution {
    
public:int rob(vector<int>& nums) {
    if (nums.empty()) return 0;if (nums.size()==1) return nums[0];// C++的数组切片vector<int> sub_nums1 = vector<int>(nums.begin() + 1, nums.end());vector<int> sub_nums2 = vector<int>(nums.begin(), nums.end()-1);int res1 = sub_rob(sub_nums1);int res2 = sub_rob(sub_nums2);return max(res1,res2);}int sub_rob(vector<int>& nums) {
    int n = nums.size();if (n==0) return 0;vector<int> dp(n+1,0);dp[1] = nums[0];for (int i=2;i<=n;++i){
    dp[i] = max(dp[i-1],dp[i-2]+nums[i-1]);} return dp[n];}
};