leetcode House Robber题解

最新推荐文章于 2021-05-17 11:01:03 发布

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本文探讨了一种专业小偷盗窃策略，通过动态规划算法在不连续盗窃的情况下最大化收益。文章详细解析了算法原理，提供了Java实现代码，适用于对算法和动态规划感兴趣的技术人员。

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题目描述：

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: [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: [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.

中文理解：一个专业的小偷入室盗窃，不过不可以盗连续两家的财务，不然就会触发报警设置，已知每家的财务，求可以偷取的最大财物。

解题思路：典型的动态规划的题目，递推公式为dp[i]=Math.max(dp[i-1],dp[i-2]+nums[i])，且dp[0]=nums[0]，dp[1]=Math.max(nums[0],nums[1])。

代码（java）：

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