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

本題是一道動態規劃題目，如果用 A[i] 表示第 i 個房子所隱藏的金額；用 dp[i] 表示劫匪到第 i 個房子時所能搶到的最大金額，則動態轉移方程爲：
$dp[i] = max ( dp[i-2] + A[i] , dp[i-1] )$
經過分析可發現，dp這個數組我們每次只需要用兩個元素的值，爲了節省空間，可以使用滾動數組來處理。

Java實現

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實現

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
    }
}