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.

解析：打家劫舍问题，意思是说一个有经验的劫匪要抢劫路边的房子，每个房子有不等的财产，唯一的限制就是不能同时抢劫相邻的两个房子，这样会触发警报。可以看出来这是一个动态规划的问题，假设dp[i]表示表示到当前位置i所打劫到的最大值，则nums[i]有两种选择，如果选择第i个房子，则第i-1个房子不能选，此时dp[i]=dp[i-1]+nums[i]；如果nums[i]不选择，则dp[i]=dp[i-1]。具体代码如下：

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