198. House Robber

Description:

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that 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: [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.

Example 2:

Input: [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
             Total amount you can rob = 2 + 9 + 1 = 12.

 思路：

图片来源：LeetCode 198. House Robber 递归->记忆->动态规划 ->新思路

解法一：递归法

时间复杂度和空间复杂度都太高，没有AC，直接Runtime Error，StackOverflowError

public class House_Robber {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		int[]  arr = {1, 2, 3, 1};
		System.out.println(rob(arr));
	}
	
    public static int rob(int[] nums) {
        return solve(nums, nums.length-1);
     }
    
	public static int solve(int[] nums, int n) {
		if(n == 0) {
			return nums[0];
		}else if (n == 1) {
			return Math.max(nums[0], nums[1]);
		}else {
		   int A = solve(nums, n-2) + nums[n];
		   int B = solve(nums, n-1);
		   return Math.max(A, B);
		}
	}
}

方法二：记忆化优化方法

时间复杂度太高，没有AC，直接Time Limit Exceeded

import java.util.Arrays;

public class House_Robber {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		int[]  arr = {1, 2, 3, 1};
		System.out.println(rob(arr));
	}
	
    public static int rob(int[] nums) {
    	if(nums.length == 0) {
    		return 0;
    	}else {
    		 return solve1(nums, nums.length-1);
		}
     }
	
	public static int solve1(int[] nums, int n) {
		int[] memo = new int[n+1];
		Arrays.fill(memo, -1);
		if(n == 0) {
			return nums[0];
		}else if(n == 1) {
			return Math.max(nums[0], nums[1]);
		}else {
			if(memo[n] == -1) {
				int A = solve1(nums, n-2) + nums[n];
				int B = solve1(nums, n-1);
				memo[n] = Math.max(A, B);
			}
		}
		return memo[n];
		
	}
}

方法三：动态规划方法

已经AC的代码

import java.util.Arrays;

public class House_Robber {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		int[]  arr = {1, 2, 3, 1};
		System.out.println(rob(arr));
	}
	
    public static int rob(int[] nums) {
    	if(nums.length == 0) {
    		return 0;
    	}else {
    		 return solve1(nums, nums.length-1);
		}
     }
    
    //动态规划实现
    public static int solve1(int[] nums, int n) {
    	int[] memo = new int[n+1];
    	Arrays.fill(memo, -1);
    	if(n == 0) {
    		return nums[0];
    	}else if (n == 1) {
			return Math.max(nums[0], nums[1]);
		}else {
			memo[0] = nums[0];
			memo[1] = Math.max(nums[0], nums[1]);
	    	for(int j=2; j<=n; j++) {
	    		if(memo[j] == -1) {
		    		int A = memo[j-2] + nums[j];
		    		int B = memo[j-1];
		    		memo[j] = Math.max(A, B);
	    		}
	    	}
		}
    	return memo[n];
    }

}

Reference：

https://blog.****.net/wys2011101169/article/details/73520653