[LeetCode] House Robber I II

最新推荐文章于 2018-11-04 20:07:33 发布

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最新推荐文章于 2018-11-04 20:07:33 发布

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原文链接：https://segmentfault.com/a/1190000004513734

本文详细探讨了两个经典计算机科学问题：House Robber I 和 House Robber II。这两个问题是动态规划的经典案例，涉及如何在不触发报警的情况下从一系列房屋中窃取最大金额。文章提供了详细的解决方案，包括边界条件的处理和代码实现。

House Robber I

Problem

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

Given [3, 8, 4], return 8.

Note

注意对边界条件的判断，是否非空，是否长度为1.

Solution

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

Update 2018-9

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

House Robber II

Problem

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, 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: [2,3,2]
Output: 3
Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2),
             because they are adjacent houses.

Example 2:

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.

Solution

class Solution {
    public int rob(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        int n = nums.length;
        if (n == 1) return nums[0];
        int[] dp = new int[n+1];
        
        //rob head
        dp[0] = 0;
        dp[1] = nums[0];
        for (int i = 2; i < n; i++) {
            dp[i] = Math.max(dp[i-1], dp[i-2]+nums[i-1]);
        }
        //save head-robbed result to temp
        int temp = dp[n-1];
        
        //not rob head
        dp[0] = 0;
        dp[1] = 0;
        for (int i = 2; i <= n; i++) {
            dp[i] = Math.max(dp[i-1], dp[i-2]+nums[i-1]);
        }
        
        //return the larger of tail-robbed and head-robbed
        return Math.max(dp[n], temp);
    }
}