198. House Robbery (Medium)

Ideas:

Every time there is an interval, the result of each time depends on the result of the previous time, I think of using dynamic programming

To meet the conditions:

1) Optimal sub-problem: the final optimal solution is generated by each sub-optimal solution

2) Overlapping sub-problems: To calculate the current solution, you have to repeatedly calculate the previous

 

Code:

class Solution {
    public int rob(int[] nums) {
		int n=nums.length;
		if(n==0) return 0;
		if(n==1) return nums[0];
		
		int[] dp=new int[n+1];

                //规律：每次初始化前2个，0和1
		dp[0]=nums[0];
		//注意不是dp[1]=nums[1]，因为还有nums[0]
		dp[1]=Math.max(dp[0],nums[1]);
		
		for(int i=2;i<n;i++){
			//若不选i，则选dp[i-1]
			//若选i，则选dp[i-2]+nums[i]
			//比较两者哪个更大
			dp[i]=Math.max(dp[i-2]+nums[i],dp[i-1]);
			
			
		}
                //因为有n个，但下标n-1才是第n个
		return dp[n-1];
    }
}

 

break down:

1) Regularity: the first 2 of each initialization, dp[0] and dp[1]

 

2) The last return is dp[n-1], because there are n, but the subscript n-1 is the nth

 

3) Satisfy the 2 conditions of dynamic programming:

       i) Optimal sub-problem: The final optimal solution is generated by each sub-optimal solution

       ii) Overlapping sub-problems: To calculate the current solution, you have to repeatedly calculate the previous

 

4) To prevent subscript overflow, when dp is declared, the length is added 1 (n+1) on the basis of n

 

5) Belongs to the second dp form:

The current value depends on all the previously calculated values