198 House Robber 打家劫舍

Description:
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:

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.

題目描述:
你是一個專業的小偷，計劃偷竊沿街的房屋。每間房內都藏有一定的現金，影響你偷竊的唯一制約因素就是相鄰的房屋裝有相互連通的防盜系統，如果兩間相鄰的房屋在同一晚上被小偷闖入，系統會自動報警。

給定一個代表每個房屋存放金額的非負整數數組，計算你在不觸動警報裝置的情況下，能夠偷竊到的最高金額。

示例：

示例 1:
輸入: [1,2,3,1]
輸出: 4
解釋: 偷竊 1 號房屋 (金額 = 1) ，然后偷竊 3 號房屋 (金額 = 3)。
偷竊到的最高金額 = 1 + 3 = 4 。

示例 2:
輸入: [2,7,9,3,1]
輸出: 12
解釋: 偷竊 1 號房屋 (金額 = 2), 偷竊 3 號房屋 (金額 = 9)，接著偷竊 5 號房屋 (金額 = 1)。
偷竊到的最高金額 = 2 + 9 + 1 = 12 。

思路:

動態規劃
狀態轉移公式為: f(n) = max[f(n - 1), f(n - 2) + an]
如果采用數組保存, 空間復雜度為O(n)(Java代碼, 自頂向下/C++代碼, 自底向上)
從狀態轉移公式可以看出, 實際上用到的只有 2個變量, 可以將空間進一步壓縮
時間復雜度O(n), 空間復雜度O(1)

代碼:
C++:

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

Java:

class Solution {
    private int[] memo;
    
    public int rob(int[] nums) {
        memo = new int[nums.length];
        Arrays.fill(memo, -1);
        return tryRob(nums, 0);
    }

    private int tryRob(int[] nums, int index) {
        if (index >= nums.length) {
            return 0;
        }
        if (memo[index] != -1) {
            return memo[index];
        }
        int result = 0;
        for (int i = index; i < nums.length; i++) {
            result = Math.max(result, nums[i] + tryRob(nums, i + 2));
        }
        memo[index] = result;
        return result;
    }
}

Python:

class Solution:
    def rob(self, nums: List[int]) -> int:
        pre, result = 0, 0
        for num in nums:
            pre, result = result, max(num + pre, result)
        return result