House Rober
来源:互联网 发布:阿里云首席科学家 年薪 编辑:程序博客网 时间:2024/05/17 03:03
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.
我是做动态规划的一道Medium题的时候返回来做的这道简单题,主要就是熟悉动态规划,本人对动态规划算法用的很少,所以集中练练动态规划的题。
首先这道题满足动态规划算法的要求,一个是最优子结构性质,二是子问题重叠性质,所以可以使用动态规划来解决问题。
其次使用一个数组dp来保存在每个房子处应该抢到钱数的最大值。第i个房子处有两种情况,可以是dp[i-1]或者是dp[i-2]+nums[i],所以每个房子处的dp[i]=max(dp[i-1],dp[i-2]+nums[i]);
具体代码如下:
int rob(vector<int>& nums)
{
if(nums.size()<=1) return nums.size()==0?0:nums[0];
vector<int> dp;
dp.push_back(nums[0]);
dp.push_back(nums[0]>=nums[1]?nums[0]:nums[1]);
for(int i=2;i<nums.size();i++)
{
dp.push_back(max(dp[i-2]+nums[i],dp[i-1]));
}
return dp.back();
}
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