[2018-12-02] [LeetCode-Week13] 198. House Robber 动态规划

https://leetcode.com/problems/house-robber/

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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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d[i] = 前 i 个房子能偷的最多的钱数
状态转移方程：d[i] = max(d[i-1], d[i-2] + nums[i])
两种情况对应是否抢第 i 家
初始条件：d[0] = nums[0], d[1] = max(d[0], nums[1])

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class Solution {
public:
    int rob(vector& nums) {
        int n = nums.size();
        if (n == 0) return 0;
        int d_2 = nums[0];
        int d_1;
        if (n > 1) {
            d_1 = max(d_2, nums[1]);
        } else {
            d_1 = 0;
        }
        int d = max(d_1, d_2);
        for (int i = 2; i < n; i++) {
            d = max(d_1, d_2 + nums[i]);
            d_2 = d_1;
            d_1 = d;
        }
        return d;
    }
    
};