House Robber
Description of problem
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 systems connected and it will automatically contact the police if two adjacent houses were broken into on the same night.
Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.
Example 1:
Input: nums = [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: nums = [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.
Constraints:
1 <= nums.length <= 1000 <= nums[i] <= 400
Problem analysis
The problem is an classical dynamic planning problem. So we just need to figure out its state transition equation; after that, we can implement it with code according to state transition equation.
Now we will analogy its state transition equation. We can first get the restrition of problem: can not rob the two adjacent house, otherwise, the security systems will be activated. So, if we have robbed the i-th house, we can not rob the (i+1)-th house. According to the rule, we can get the state transition equation: f[i]=max(f[i-1],f[i-2]+nums[i]).
Implementation Code
class Solution {
public:
// can not rob the two adjacent houses on the same night
// f[i]=max(f[i-1],f[i-2]+nums[i])
int rob(vector<int>& nums) {
int n=nums.size();
if(n==1)
return nums[0];
int f[n];
f[0]=nums[0];f[1]=max(f[0],nums[1]);
for(int i=2;i<n;i++){
f[i]=max(f[i-1],f[i-2]+nums[i]);
}
return f[n-1];
}
};