[LeetCode ] House Robber

最新推荐文章于 2024-01-18 14:02:27 发布

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最新推荐文章于 2024-01-18 14:02:27 发布

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本文探讨了一个经典的计算机科学问题——强盗抢劫问题，即如何在不连续抢劫相邻房屋的情况下，最大化所获金额。通过使用动态规划方法，文章详细解释了如何计算并返回在给定条件下可抢夺的最大金额。

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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.

题意：强盗抢劫，不能抢相邻的两家，问最多能抢多少钱？

思路：线性dp，dp[i][0]表示不抢第i家的最多抢多少钱，dp[i][1]表示抢第i家最多抢多少钱。

dp[i][0]=max( dp[i-1][0],dp[i-1][1] )

dp[i][1]=nums[i] + max( dp[i-1][0],dp[i-2][1] )

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