LeetCode 213. House Robber II(动态规划)

213. House Robber II

Medium

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.

题意

给定n个正整数组成的数组,不能同时取相邻两个元素,也不能同时取首元素和尾元素,求能够取到的最大的和。

思路

首先将循环问题转化为两个非循环问题求解(一个非循环问题可以使用首元素不能使用尾元素,另一个非循环问题可以使用尾元素不能使用首元素)。对于每个非循环问题,用动态规划求解。具体来说,维护两个值pre和cur, 分别代表nums[0:i-1]和nums[0:i]的子问题的和,动态规划递推式为
dp[0:i+1] = max(dp[0:i-1] + nums[i], dp[0:i])

代码

class Solution {
    public int rob(int[] nums) {
        int n = nums.length;
        if (n == 0) {
            return 0;
        }
        if (n == 1) {
            return nums[0];
        }
        return Math.max(myrob(nums, 0, n-2), myrob(nums, 1, n-1));
    }
    
    private int myrob(int[] nums, int begin, int end) {
        if (begin == end) {
            return nums[begin];
        }
        int pre = nums[begin], cur = Math.max(nums[begin], nums[begin+1]);
        for (int i = begin+2; i <= end; ++i) {
            int tmpcur = cur;
            cur = Math.max(pre + nums[i], cur);
            pre = tmpcur;
        }
        return cur;
    }
}

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