LeetCode 198. House Robber--动态规划--C++，Java，Python解法

此题链接：House Robber - LeetCode
LeetCode 动态规划（Dynamic programming）系列题目:LeetCode 动态规划（Dynamic programming）系列题目

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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.

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这道题目是标准的动态规划，可以用递归或者迭代。递归做法比较简单，但是很容易太深，迭代做法即可。
Python解法如下：

class Solution:
    def rob(self, nums: List[int]) -> int:
        count=0
        if nums==[]:
            return 0
        l=[0 for i in range(0,len(nums))]
        for i in range(0,len(nums)):
            if i==0:
                l[0]=nums[0]
            elif i==1:
                l[1]=max(l[0],nums[1])
            else:
                l[i]=max(l[i-2]+nums[i],l[i-1])
        return l[-1]

Java解法如下：

class Solution {
  public int rob(int[] num) {
    int prevMax = 0;
    int currMax = 0;
    for (int x : num) {
        int temp = currMax;
        currMax = Math.max(prevMax + x, currMax);
        prevMax = temp;
    }
    return currMax;
}
}

C++解法如下：

class Solution {
public:
    int rob(vector<int>& nums) {
        const int n = nums.size();
        if (n == 0) return 0;
        if (n == 1) return nums[0];
        if (n == 2) return max(nums[0], nums[1]);
        vector<int> f(n, 0);
        f[0] = nums[0];
        f[1] = max(nums[0], nums[1]);
        for (int i = 2; i < n; ++i)
            f[i] = max(f[i-2] + nums[i], f[i-1]);
        return f[n-1];
    }
};