Home raiding II (leetcode-213)

raid homes and plunder houses Ⅱ（LeetCode-213）

subject

You are a professional thief , Plan to steal houses along the street , There is a certain amount of cash in every room . All the houses in this place are Make a circle , This means that the first house and the last house are next to each other . meanwhile , Adjacent houses are equipped with interconnected anti-theft system , If two adjacent houses are broken into by thieves on the same night , The system will automatically alarm .

Given an array of non negative integers representing the storage amount of each house , Count you Without triggering the alarm device , The maximum amount you can steal tonight .

Example 1：

 Input ：nums = [2,3,2]
 Output ：3
 explain ： You can't steal first  1  House No （ amount of money  = 2）, And then steal  3  House No （ amount of money  = 2）,  Because they are next to each other .

Example 2：

 Input ：nums = [1,2,3,1]
 Output ：4
 explain ： You can steal first  1  House No （ amount of money  = 1）, And then steal  3  House No （ amount of money  = 3）.
      Maximum amount stolen  = 1 + 3 = 4 .

Example 3：

 Input ：nums = [1,2,3]
 Output ：3

Tips ：

• 1 <= nums.length <= 100
• 0 <= nums[i] <= 1000

Ideas

When the number of rooms is no more than two , There is no need to consider the beginning and end

More than two hours , Because you can't steal both rooms at the same time , So we can consider it in two cases

1. Don't steal the last case ： The scope of theft $nums[0:n-2]$
2. Don't steal the first case ： The scope of theft $nums[1:n-1]$

The greater of the two

Code display

class Solution
{
    
public:
    int rob(vector<int> &nums)
    {
    
        int n = nums.size();
        if (n == 1)
        {
    
            return nums[0];
        }
        int left = robRange(nums, 0, n - 2);
        int right = robRange(nums, 1, n - 1);
        return max(left, right);
    }
    int robRange(vector<int> &nums, int start, int end)
    {
    
        if (start == end)
        {
    
            return nums[start];
        }
        vector<int> dp(nums.size());
        dp[start] = nums[start];
        dp[start + 1] = max(nums[start], nums[start + 1]);
        for (int i = start + 2; i <= end; i++)
        {
    
            dp[i] = max(dp[i - 1], dp[i - 2] + nums[i]);
        }
        return dp[end];
    }
};