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.

 

思路：这题的状态转移方程还是比较好找的，然而我转换代码的能力还是太弱，或者说dp题做的还不够多，费了好多时间。。。

方程dp[i] = max{nums[i] + (i -2 < 0 ? 0 : dp[i-2]), dp[i-1]}

1 class Solution { 2 public: 3     int rob(vector
     
      & nums) { 4         if (!nums.size()) 5             return 0; 6         int dp[nums.size()]; dp[0] = nums[0]; 7         for (int i = 1; i != nums.size(); ++i) 8             dp[i] = max(nums[i] + (i-2 < 0 ? 0 : dp[i-2]), dp[i-1]); 9         return dp[nums.size()-1];10     }11 };