(DP 背包) leetcode 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.

Example 1:

Input: [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
             Total amount you can rob = 1 + 3 = 4.

Example 2:

Input: [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
             Total amount you can rob = 2 + 9 + 1 = 12.

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这个应该是背包问题吧。本质是求数组不相邻元素最大和。有偷第i个和不偷第i个的两种不同方案。

这个题使用dp求解，状态转换式为：dp[i] = max(num[i] + dp[i - 2], dp[i - 1]) ，其中dp[0] = nums[0], dp[1] = max(dp[0],dp[1])；

参考链接：http://www.cnblogs.com/grandyang/p/4383632.html

C++代码：

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

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.

Example 1:

Input: [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
             Total amount you can rob = 1 + 3 = 4.

Example 2:

Input: [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
             Total amount you can rob = 2 + 9 + 1 = 12.