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翻译
你是一个专业强盗，并计划沿街去盗窃每一个住户。
每个房子都有一定量的现金，阻止你盗窃的唯一阻碍是相邻的两个房子之间有安全系统。
一旦这两个房子同时被盗窃，系统就会自动联系警察。
给定一系列非负整数代表每个房子的金钱，
求出再不惊动警察的情况下能盗窃到的最大值。原文
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.分析
典型的动态规划问题。
int robIter(vector<int>& money, int c, int dp1, int dp2) {
if (c >= money.size()) return dp2;
else return robIter(money, c, dp2, max(dp1 + money[c++], dp2));
}
int rob(vector<int>& nums) {
return robIter(nums, 0, 0, 0);
}上面写的可能太简洁，这样或许方面理解一些：
int robIter(vector<int>& money, int c, int dp1, int dp2) {
if (c >= money.size())
return dp2;
else {
dp1 = max(dp1 + money[c++], dp2);
return robIter(money, c, dp2, dp1);
}
}
int rob(vector<int>& nums) {
return robIter(nums, 0, 0, 0);
}代码
class Solution {
public:
int robIter(vector<int>& money, int c, int dp1, int dp2) {
if (c >= money.size()) return dp2;
else return robIter(money, c, dp2, max(dp1 + money[c++], dp2));
}
int rob(vector<int>& nums) {
return robIter(nums, 0, 0, 0);
}
};
