198. House-Robber

difficulty: Easy

section pre{ background-color: #eee; border: 1px solid #ddd; padding:10px; border-radius: 5px; }

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: 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:

• 0 <= nums.length <= 100

• 0 <= nums[i] <= 400

Method One

class Solution {
    public int rob(int[] nums) {
        if( nums.length < 1 ) {
            return 0;
        }
        int[] dp = new int[nums.length];
        for(int i = 0; i < nums.length; i++ ) {
            int prePre = i - 2 < 0 ? 0 : dp[i - 2];
            int pre = i - 1 < 0 ? 0 : dp[ i - 1 ];
            dp[i] = Math.max(prePre + nums[i], pre);
        }
        return dp[nums.length - 1];
    }
}
​
/**
index 0 1 2 3 4
value 3 6 9 2 8
dp[0] = nums[0];
dp[1] = Math.max(  nums[1], dp[0] );
dp[2] = Math.max(dp[0] + nums[2], dp[1]);
...
dp[i] = Math.max( dp[i - 2] + nums[i], dp[ i - 1 ] );
*/
​
​

当然很轻松就可以化成 O(1) 的。

class Solution {
    public int rob(int[] nums) {
        if( nums.length < 1 ) {
            return 0;
        }
        int prePre = 0;
        int pre = 0;
        for(int i = 0; i < nums.length; i++ ) {
            int temp = pre;
            pre = Math.max( prePre + nums[i], pre);
            prePre = temp;
        }
        return pre;
    }
}

2022 写的答案。 注意一个很重要的观察是，对于house robber来说，这个动态规划的结果永远有 dp[x] >= dp[x - 1]. 因此在skip当前房子的情况下，只需要看 dp[x - 1]就好了。

class Solution {
    public int rob(int[] nums) {
        if(nums.length == 0) return 0;
        if(nums.length == 1) return nums[0];
        // RobCurrent = check n - 2 + curVal
        // SkipCurrent = check n - 2 and n - 1
        // max(rob, skip)
        int[] dp = new int[nums.length + 1];
        dp[0] = 0;
        dp[1] = nums[0];
        for(int i = 2; i <= nums.length; i++){
            int robCurHouse = dp[i - 2] + nums[i - 1];
            // int skip = Math.max(dp[i - 1], dp[i - 2]);
            int skipCurHouse = dp[i - 1]; // since dp[i - 1] >= dp [i - 2];
            dp[i] = Math.max(robCurHouse, skipCurHouse);
        }
        return dp[nums.length];
    }
}