【算法作业12】LeetCode 198. House Robber

198. House Robber

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.

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题解：

这是一道动态规划的题目。题意是去盗窃一条路上的房子，如果盗窃两个连在一起的房子则会引发警报，因此只能盗窃不连续的房子，求能够盗窃到的最大金额数目。

题目给了一个vector<int> nums来存放每个房子的价值，我们再假设盗窃第i个和它之前的房子获得的最大金额为amount[i]。由于只能盗窃不相邻的房子，因此在决定一间房子i是否盗窃的时候只有两种方案，第一种是盗窃这个房子，那么此时获得的金钱就是nums[i] + amount[i
- 2]；另一种方案是不盗窃这个房子，那么此时获得的金钱就是amount[i - 1]。遍历整个vector<int> nums，对每个房子选取获得最大金钱值的方案，最后就能获得最终结果。

代码：

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