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.

方法1：

public class Leetcode_198_HouseRobber_DP1 {
    public static void main(String[] args) {
        Leetcode_198_HouseRobber_DP1 object = new Leetcode_198_HouseRobber_DP1();
        int[] nums = new int[]{2, 7, 9, 3, 1};
//        int[] nums = new int[]{};
        System.out.println(object.rob(nums));
    }

    public int rob(int[] nums) {
        if (nums.length ==  0) {//[]不等於null，需要用長度判斷
            return 0;
        }
        if (nums != null && nums.length == 1) {
            return nums[0];
        }
        if (nums.length == 2) {
            Math.max(nums[0], nums[1]);
        }
        int[] dp = new int[nums.length];
        dp[0] = nums[0];
        dp[1] = Math.max(nums[0], nums[1]);//因為nums[0]與nums[1]均可以與nums[3]組合，所以取兩者的最大值
        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];
    }
}
 

輸出12

Runtime: 4 ms, faster than 41.21% of Java online submissions for House Robber.

end