Looting (leetcode-198)

raid homes and plunder houses （LeetCode-198）

subject

You are a professional thief , Plan to steal houses along the street . There is a certain amount of cash in every room , The only restriction on your theft is that adjacent houses are equipped with interconnected anti-theft systems , 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 , The maximum amount that can be stolen overnight .

Example 1：

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

Example 2：

 Input ：[2,7,9,3,1]
 Output ：12
 explain ： Steal  1  House No  ( amount of money  = 2),  Steal  3  House No  ( amount of money  = 9), Then steal  5  House No  ( amount of money  = 1).
      Maximum amount stolen  = 2 + 9 + 1 = 12 .

Tips ：

• 1 <= nums.length <= 100
• 0 <= nums[i] <= 400

Ideas

Five steps

1. dp[i] meaning

   • Before stealing i Houses （ Including house i） The maximum amount that can be obtained

2. The recursive formula

   • $$dp[i]=max(dp[i-1],dp[i-2]+nums[i])$$

3. Array initialization

   • dp[0]=0 dp[1]=nums[0]

4. traversal order

   • Before and after

Code

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