198. House Robber

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本文介绍了一种使用动态规划解决的专业窃贼问题。该问题要求窃贼不能连续抢劫相邻的房屋，需要通过动态规划算法来计算出不触发警报情况下能获得的最大金额。

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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.

这道题就是采用动态规划的方法，要隔着取数求n个数的最大和,其实就是先取n-1个数的最大和，要取n-1个数的最大和，其实是得先取n-2个数的最大和。依次递归。然后是每次比较n-1个数的最大和的大小和隔着的那个数的大小，求得最大的那个数。

代码如下：

class Solution {
public:
    int rob(vector<int>& nums) {
        if(nums.size()==0)return 0;

        int len=nums.size();
        int tmp[len];

        if(nums.size()==1){
            return nums[0];
        }

        if(nums.size()==2){
            if(nums[0]>nums[1])return nums[0];
            else return nums[1];
        }

        for(int i=0;i<len;i++){
            if(i==0)tmp[i]=nums[i];
            else if(i==1){
                tmp[i]=max(tmp[i-1],nums[i]);
            }
            else{
                tmp[i]=max(tmp[i-1],tmp[i-2]+nums[i]);
            }
        }
        return tmp[len-1]>tmp[len-2]?tmp[len-1]:tmp[len-2];

    }
};