#392 House Robber

最新推荐文章于 2021-09-27 21:36:25 发布

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最新推荐文章于 2021-09-27 21:36:25 发布

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分类专栏： lintcode 文章标签： lintcode dynamic programming

本文链接：https://blog.youkuaiyun.com/haifischxia/article/details/52250729

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本文探讨了一道经典的动态规划问题——打家劫舍，旨在寻找从一系列房屋中盗窃而不触发报警系统的最大收益。文章提供了一个简洁高效的解决方案，通过递推公式实现了O(n)时间复杂度与O(1)空间复杂度。

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题目描述：

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?

Yes

Example

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

Challenge

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

题目思路：

这题倒是明显的dp题，假设dp[i]是rob第i个house所得的最大金额，那么这个小偷有两种选择：不偷这个house，和偷这个house。在这两种选择中取金额最大的选项，就是dp[i]。

Mycode（AC = 513ms）：

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
        if (A.size() == 0) {
            return 0;
        }
        else if (A.size() == 1) {
            return A[0];
        }
        else if (A.size() == 2) {
            return max(A[0], A[1]);
        }
        else {
            vector<long long> dp(A.size() + 1, 0);
            dp[1] = A[0];
            dp[2] = max(A[0], A[1]);
            
            // pick the max of robbing i-th house or not robbing house i-th house
            for (int i = 3; i <= A.size(); i++) {
                dp[i] = max(dp[i - 2] + (long long)A[i - 1], dp[i - 1]);
            }
            
            return dp[A.size()];
        }
    }
};