LeetCode-in-All
198. House Robber
Medium
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
Solution
class Solution {
int rob(List<int> nums) {
// The best you can do at any given house is
// the value of this hosue + maximum values of
// what you could get at every house that is
// two away from this one.
// Expressed as pesudocode
// dp[i] = i + max(dp[i-2] ..... dp[0])
List<int> dp = List<int>.filled(nums.length, 0);
int maxValue = 0;
for (int i = 0; i < nums.length; i++) {
// i is the index we are filling the table for
// Look back at every dp before this.
// Start with the value of this house
int value = nums[i];
dp[i] = value;
for (int j = 0; j < i - 1; j++) {
dp[i] = max(dp[i], value + dp[j]);
}
maxValue = max(maxValue, dp[i]);
}
return maxValue;
}
}