House Robber
You are a professional thief 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.
Constraints:
- 1 <= nums.length <= 100
- 0 <= nums[i] <= 100
Examples:
Input: [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3). Total money you can rob = 1 + 3 = 4.
Solutions
Dynamic Programming
Create a dynamic programming table dp where dp[i] represents the maximum amount of money that can be robbed up to the i-th house. For each house, we have two options: rob the current house or not. If we rob the current house, we cannot rob the previous house, so the maximum amount of money we can rob is dp[i - 2] + nums[i]. If we do not rob the current house, the maximum amount of money we can rob is dp[i - 1]. We choose the maximum of these two options.
class Solution {
public: int rob(vector<int>& nums) {
vector<int> dp(nums.size(), 0);
dp[0] = nums[0];
for (int i = 1;
i < nums.size();
i++) {
if (i == 1) {
dp[i] = max(nums[0], nums[1]);
}
else {
dp[i] = max(dp[i - 1], dp[i - 2] + nums[i]);
}
}
return dp[nums.size() - 1];
}Follow-up:
House Robber II
