House Robber

30% Accepted

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.

Have you met this question in a real interview? Yes
Example
Given [3, 8, 4], return 8.

Challenge

• O(n) time and O(1) memory.

Tags Expand

• Dynamic Programming

O(n) time and O(n) memory.

• DP
• ret[i] = Math.max(ret[i-2]+A[i-1],ret[i-1]);
• ret[0] = 0, 必须初始化第一步

public class Solution {
    /**
     * @param A: An array of non-negative integers.
     * return: The maximum amount of money you can rob tonight
     */
    public long houseRobber(int[] A) {
        // write your code here
        if (A.length == 0) {
            return 0;
        }
        long[] ret = new long[A.length+1];
        ret[0] = 0;
        ret[1] = A[0];
        long max = ret[1];
        for (int i = 2; i <= A.length; i++) {
            ret[i] = Math.max(ret[i-2]+A[i-1],ret[i-1]);
            max = Math.max(max, ret[i]);
        }
        return max;
    }
}

O(n) time and O(1) memory.

• 其实上个版本只是用到了ret[i-2] ret[i-1]，那么只需要用两个变量去替换，然后每次循环更新两个变量的值就可以了

public class Solution {
    /**
     * @param A: An array of non-negative integers.
     * return: The maximum amount of money you can rob tonight
     */
    public long houseRobber(int[] A) {
        // write your code here
        if (A.length == 0) {
            return 0;
        }
        long first = 0;
        long second = A[0];
        long max = second;
        for (int i = 2; i <= A.length; i++) {
            max = Math.max(Math.max(first+A[i-1],second), max);
            first = second;
            second = max;
        }
        return max;
    }
}

换种写法

public class Solution {
    public int rob(int[] nums) {
        int size = nums.length;
        if (size == 0) {
            return 0;
        }
        int[] dp = new int[2];
        dp[0] = 0;
        dp[1] = nums[0];

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

        return dp[size % 2];
    }
}