Likou-198 Robbery (dp)

Likou-198 Robbery

1. Topic

198. Robbery

You are a professional thief planning to steal houses along the street. There is a certain amount of cash hidden in each room. The only restrictive factor that affects your theft is that the 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 call the police. .

Given an array of non-negative integers representing the amount stored in each house, calculate the maximum amount you can steal in one night without triggering the alarm .

Example 1:

输入：[1,2,3,1]
输出：4
解释：偷窃 1 号房屋 (金额 = 1) ，然后偷窃 3 号房屋 (金额 = 3)。
     偷窃到的最高金额 = 1 + 3 = 4 。

2. Analysis

1. topic. Seeing this topic, the first thing that comes to mind is that they cannot be adjacent , so if we want to steal ione of them, then we need to consider that the previous one i-1cannot be stolen, and i-2the other one can be stolen , so we can probably know that this is a dynamic programming problem .
2. According to the above analysis, dp[i] is the maximum amount when we steal the current i house. Then we can deduce the formula as: dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]).
3. initialization. According to the deduction formula, it is obvious that the two numbers in front of dp are needed, so dp[0] and dp[1] are what we need.
4. traverse. Start with the 2nd and work backwards.

3. Code and comments

class Solution {
    
    
    public int rob(int[] nums) {
    
    
        // 1. dp[i]表示到第i个房间能够偷窃的最高金额
        if (nums.length == 0) return 0;
        if (nums.length == 1) return nums[0];
        int[] dp = new int[nums.length + 1];
        // 2. 因为递推公式需要dp前两个数，需初始化
        dp[0] = nums[0];
        dp[1] = Math.max(nums[0], nums[1]);


        for (int i = 2; i < nums.length; i++){
    
    
            dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]);
        }

        return dp[nums.length - 1];
    }
}

4. Practice

Link to link to the topic : 198. Robbery