LeetCodse198. House Robber（动态规划）

LeetCodse198. House Robber（动态规划）

• Easy
• Accepted：302,196
• Submissions：739,457

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 1:

Input: [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
Total amount you can rob = 1 + 3 = 4.

Example 2:

Input: [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
Total amount you can rob = 2 + 9 + 1 = 12.

链接

https://leetcode.com/problems/house-robber

题解

$dp[i] = max(dp[i], nums[j] + (j + 2 < n \ ? \ dp[j + 2] : 0));​$

代码

• Runtime: 12 ms, faster than 18.25% of C++ online submissions for House Robber.
• Memory Usage: 8.6 MB, less than 100.00% of C++ online submissions for House Robber.
class Solution {
private:
vector<int> vis;
int solve(vector<int>& nums, int index)
{
if (index >= nums.size())
return 0;
if (vis[index] != -1)
return vis[index];
int res = 0;
for (int i = index; i < nums.size(); ++i)
res = max(res, nums[i] + solve(nums, i + 2));
vis[index] = res;
cout << res << endl;
return res;
}

public:
int rob(vector<int>& nums) {
vis = vector<int> (nums.size(), -1);
return solve(nums, 0);
}
};
• Runtime: 4 ms, faster than 100.00% of C++ online submissions for House Robber.
• Memory Usage: 8.6 MB, less than 100.00% of C++ online submissions for House Robber.
class Solution {
public:
int rob(vector<int>& nums) {
int n = nums.size();
if (n == 0)
return 0;
vector<int> dp = vector<int> (n + 1, -1);
dp[n - 1] = nums[n - 1];
for (int i = n - 2; i >= 0; --i)
{
for (int j = i; j < n; ++j)
dp[i] = max(dp[i], nums[j] + (j + 2 < n ? dp[j + 2] : 0));
}
return dp[0];
}
};
• Runtime: 3 ms, faster than 50.37% of C++ online submissions for House Robber.
• Memory Usage: 7.7 MB, less than 45.57% of C++ online submissions for House Robber.
class Solution
{
public:
int rob(vector<int> &nums)
{
n = nums.size();
if (n == 1)
return nums[0];
if (n == 2)
return max(nums[0], nums[1]);
vector<int> dp(nums);
dp[1] = max(nums[0], nums[1]);
for (int i = 2; i < n; ++i)
dp[i] = max(dp[i] + dp[i-2], dp[i-1]);
return dp[n - 1];
}
private:
int n = 0;
};

The end.
2019年4月3日 星期三