House Robber

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Description:

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.

Analyse:

Considering Dynamic Planning, the transfer function is dp[i] = max(dp[i-2] + num[i], dp[i-1]).

Solution:

 1 class Solution {
 2 public:
 3     int rob(vector<int> &num) {
 4         int length = num.size();
 5         if(length == 0) return 0;
 6         else if(length == 1) return num[0];
 7         else{
 8             vector<int> dp = {num[0], max(num[1], num[0])};
 9             for(int i = 2 ; i < length; i++){
10                 dp.push_back(max(dp[i-2] + num[i], dp[i-1]));
11             }
12             return dp.back();
13         }
14     }
15 };

House Robber

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原文地址：http://www.cnblogs.com/amazingzoe/p/4403454.html