House Robber

最新推荐文章于 2021-08-05 11:16:28 发布

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本文提供了一种动态规划解决方案，用于在不触发相邻房屋警报系统的情况下，最大化抢劫财富。通过使用动态公式dp[i]=Math.max(nums[i]+dp[i-2],dp[i-1])，我们可以计算出在给定房屋列表中，不触发警报系统时能获得的最大财富。以列表[2,3,4,1]为例，我们逐步计算dp数组，最终得出最大可抢夺财富为6。

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House Robber

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.

Credits:
Special thanks to @ifanchu for adding this problem and creating all test cases. Also thanks to @ts for adding additional test cases.

Solution

This is a typical dynamic programming program.

Supposed that dp[i] represents that the maximum amount of money the robber can rob when he reaches the n house.

The dynamic formula is dp[i] = Math.max(nums[i] + dp[i - 2], dp[i - 1])

The base case is dp[0] = nums[0], dp[1] = Math.max(dp[0], nums[1])

For example, [2,3,4,1]

dp[0] = 2,

dp[1] = 3,

dp[2] = Math.max(dp[1], nums[2] + dp[0]) = 6,

dp[3] = Math.max(dp[2], nums[3] + dp[1]) = 6.

public class Solution {
    public int rob(int[] nums) {
        if (nums == null || nums.length == 0)
            return 0;
        
        int[] dp = new int[nums.length];
        
        
        dp[0] = nums[0];
        
        for (int i = 1; i < nums.length; i++) {
            if (i == 1) {
                dp[i] = Math.max(nums[0],nums[1]);
            } else {
            dp[i] = Math.max(dp[i - 1], nums[i] + dp[i - 2]);
            }
        }
        
        
        return dp[nums.length - 1];
    }
}