lintcode: House Robber

最新推荐文章于 2020-05-18 13:56:57 发布

martin_liang

最新推荐文章于 2020-05-18 13:56:57 发布

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分类专栏： C++/C 算法 lintcode

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本文探讨了一个经典的动态规划问题——打家劫舍。在这个问题中，你作为一个专业的大盗，计划抢劫沿街的房屋。每栋房子都有一定数量的钱财，但相邻的房子有联动的安全系统，如果两栋相邻的房子在同一晚被抢劫，则会自动报警。文章提供了如何通过动态规划的方法找到最大的可抢劫金额，并附带了一个 C++ 的实现案例。

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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.

Have you met this question in a real interview?

Example

Given [3, 8, 4], return 8.

Challenge

O(n) time and O(1) memory.

class Solution {
public:
    /**
     * @param A: An array of non-negative integers.
     * return: The maximum amount of money you can rob tonight
     */

     
    long long houseRobber(vector<int> A) {
        // write your code here
        long long prepreValue;
        long long preValue;
        long long curValue;
        
        if (A.size() == 0)
            return 0;
            
        if (A.size() == 1)
            return A[0];
            
        if (A.size() == 2)
            return max(A[0], A[1]);
            
        prepreValue = A[0];
        preValue    = A[1];
        
        for (int i=2; i<A.size(); i++) {
            curValue = max(prepreValue+A[i], preValue);
            
            prepreValue = preValue;
            preValue    = curValue;
        }
        return curValue;
    }
};