Leetcode题目：House Robber

题目：

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.

题目解答：这是个很典型的动态规划的问题。

在碰到当前这家时，要不要抢劫，取决于其抢前一家基础上所能获得的最大值和抢前两家加上当前这一家的钱的最大值的比较结果。如果，抢前一家获得了最大，那么就不要抢这家了；否则，就需要累加它。

其实说起来，就是在一个非负数组中，不能连续取数，求出可以取得的总和的最大值。

代码如下：

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