题目描述

[MEDIUM]

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 systems connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.

Example 1:

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Input: nums = [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:

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Input: nums = [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.

思路

dp[i] = max(dp[i-2] + nums[i], dp[i-1]), dp[i] is the max amount of money you can rob from the first i houses. And it can be derived from the last house or the second last house with the current house.

代码

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class Solution {
public:
    vector<int> dp;

    int rob(vector<int>& nums) {
        if(nums.empty()) return 0;
        if(nums.size() == 1) return nums[0];

        //init dp
        dp.push_back(nums[0]);
        int max1 = nums[0] > nums[1] ? nums[0] : nums[1];
        dp.push_back(max1);

        for(int i = 2; i < nums.size(); i++){
            int newMax = max(dp[i- 2] + nums[i], dp[i -1]);
            dp.push_back(newMax);
        }

        return dp[nums.size()- 1];
    }
};