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
Given [3, 8, 4], return 8.
Challenge
O(n) time and O(1) memory.
LeetCode上的原题，请参见我之前的博客House Robber。
解法一：
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) { if (A.size() <= 1) return A.empty() ? 0 : A[0]; vector<long long> dp{A[0], max(A[0], A[1])}; for (int i = 2; i < A.size(); ++i) { dp.push_back(max(dp[i - 2] + A[i], dp[i - 1])); } return dp.back(); } };
解法二：
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) { long long a = 0, b = 0; for (int i = 0; i < A.size(); ++i) { if (i % 2 == 0) { a += A[i]; a = max(a, b); } else { b += A[i]; b = max(a, b); } } return max(a, b); } };
解法三：
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) { long long a = 0, b = 0; for (int i = 0; i < A.size(); ++i) { long long m = a, n = b; a = n + A[i]; b = max(m, n); } return max(a, b); } };
本文转自博客园Grandyang的博客，原文链接：打家劫舍[LintCode] House Robber ，如需转载请自行联系原博主。
