題目

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.

題意

你是一個專業的強盜，你計劃在一條路上搶劫房屋。每一個房屋藏著一些錢。因為相鄰房屋之間有安全系統相連，這個安全系統又與警察相連，所以你不能搶劫相鄰的房屋在同一個晚上。
給你一列非負整數代表每一個房屋的藏的錢數，求出在沒有驚擾警察的情況下，能搶劫到的最多的錢數。

題解

不能取相鄰元素，動態規劃的思想是將大問題轉換為逐個的小問題。

假設：

1. 只有1個房屋nums[0]，最大收益為dp[0] = nums[0];
2. 有2個房屋nums[0], nums[1], 不能同時取，最大收益為dp[1] = max(nums[0], nums[1]);
3. 有3個房屋，有兩種取法，取nums[1],或者取nums[0]和nums[2].即 dp[2] = max(nums[1], nums[0] + nums[2]);
4. 故可推測出動態轉換方程為：dp[i] = max(nums[i] + dp[i-2], dp[i-1]);

C++程式碼

class Solution {
public:
    int rob(vector<int>& nums) {
        int len = nums.size();
        int dp[len+5] = {0};
        if(len==0)
            return 0;
        if(len==1)
            return nums[0];
        if(len==2)
            return max(nums[0], nums[1]);

        dp[0] = nums[0];
        dp[1] = max(nums[0], nums[1]);
        for(int i=2; i<len; i++)
        {
            dp[i] = max(nums[i] + dp[i-2], dp[i-1]);
        }
        return dp[len-1];
    }
};

python程式碼

class Solution(object):
    def rob(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        n =  len(nums);
        dp = [0 for x in range(n)];
        if n==0:
            return 0
        if n==1:
            return nums[0]
        if n==2:
            return max(nums[0], nums[1])
        dp[0] = nums[0]
        dp[1] = max(nums[0], nums[1])
        for i in range(2, n):
            dp[i] = max(nums[i] + dp[i-2], dp[i-1])
        return dp[n-1]