Leetcode198. House Robber

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探讨了在不触动相邻房屋警报系统的情况下，如何通过动态规划算法计算出从一系列房屋中能偷盗的最大金额。示例展示了算法的运行过程及结果。

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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.

动态规划，0ms

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