LeetCode-198: House Robber（打家劫舍）

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

本题是一道动态规划题目，如果用 A[i] 表示第 i 个房子所隐藏的金额；用 dp[i] 表示劫匪到第 i 个房子时所能抢到的最大金额，则动态转移方程为：
$dp[i] = max ( dp[i-2] + A[i] , dp[i-1] )$
经过分析可发现，dp这个数组我们每次只需要用两个元素的值，为了节省空间，可以使用滚动数组来处理。

Java实现

class Solution {
    public int rob(int[] nums) {
        int length = nums.length;
        if (nums == null || length == 0) {
            return 0;
        } else if (length == 1) {
            return nums[0];
        }
        int l = nums[0];
        int p = Math.max(nums[0], nums[1]);
        for (int i = 2; i < length; i++) {
            int tmp = p;
            p = Math.max(l + nums[i], p);
            l = tmp;
        }
        return p;
    }
}

Python实现

class Solution(object):
    def rob(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        length = len(nums)
        if nums == None or length == 0:
            return 0
        elif length == 1:
            return nums[0]
        
        l = nums[0]
        p = max(nums[0], nums[1])
        for i in range(2,length):
            tmp = p
            p = max(l+nums[i], p)
            l = tmp
        
        return p

Scala

object Solution {
    def rob(nums: Array[Int]): Int = {
        import scala.math._
        val length = nums.length
        if (nums == null || length == 0) {
            return 0
        } else if (length == 1) {
            return nums(0)
        }
        var l = nums(0)
        var p = max(nums(0), nums(1))
        for (i <- 2 until nums.length) {
            val tmp = p
            p = max(l + nums(i), p)
            l = tmp
        }
        return p
    }
}