[Leedcode] [JAVA] [masseur] [Dynamic Programming]

【Problem Description】

一个有名的按摩师会收到源源不断的预约请求，每个预约都可以选择接或不接。在每次预约服务之间要有休息时间，因此她不能接受相邻的预约。给定一个预约请求序列，替按摩师找到最优的预约集合（总预约时间最长），返回总的分钟数。

注意：本题相对原题稍作改动

 

示例 1：

输入： [1,2,3,1]
输出： 4
解释： 选择 1 号预约和 3 号预约，总时长 = 1 + 3 = 4。
示例 2：

输入： [2,7,9,3,1]
输出： 12
解释： 选择 1 号预约、 3 号预约和 5 号预约，总时长 = 2 + 9 + 1 = 12。
示例 3：

输入： [2,1,4,5,3,1,1,3]
输出： 12
解释： 选择 1 号预约、 3 号预约、 5 号预约和 8 号预约，总时长 = 2 + 4 + 3 + 3 = 12。

[Thinking] answer

1. Dynamic Programming to-back two-dimensional state variables to think clearly every step

Time complexity: O (N) space complexity: O (N)

1. Design status

• dp [i] [0] represents: the interval [0, i] in the reservation request accepted, and the subscript i is the maximum length of the day are not accepted reservation;
• dp [i] [1] represents: the interval [0, i] in the accepted reservation request, and the maximum duration for that day subscript i is accepts reservation.

2. State transition equation

• Today not accept reservation: yesterday or not to accept the reservation or booking last received, whichever is the maximum value, namely: dp [i] [0] = max (dp [i - 1] [0], dp [i --1] [1]);
• Today, by appointment: just do not accept an appointment from yesterday transferred from, coupled with today's often, namely: dp [i] [1] = dp [i - 1] [0] + nums [i]?.

3. Initialize
dp [0] [0] = 0 and dp [0] [1] = nums [0];

 public int massage(int[] nums) {
        int len = nums.length;
        if (len == 0) {
            return 0;
        }
        if (len == 1) {
            return nums[0];
        }

        // dp[i][0]：区间 [0, i] 里接受预约请求，并且下标为 i 的这一天不接受预约的最大时长
        // dp[i][1]：区间 [0, i] 里接受预约请求，并且下标为 i 的这一天接受预约的最大时长
        int[][] dp = new int[len][2];
        dp[0][0] = 0;
        dp[0][1] = nums[0];

        for (int i = 1; i < len; i++) {
            dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1]);
            dp[i][1] = dp[i - 1][0] + nums[i];
        }
        return Math.max(dp[len - 1][0], dp[len - 1][1]);
    }

Optimization: an array of multi-state setting line, in order to avoid extreme example will be discussed with

public class Solution {

    public int massage(int[] nums) {
        int len = nums.length;

        // dp 数组多设置一行，相应地定义就要改变，遍历的一些细节也要相应改变
        // dp[i][0]：区间 [0, i) 里接受预约请求，并且下标为 i 的这一天不接受预约的最大时长
        // dp[i][1]：区间 [0, i) 里接受预约请求，并且下标为 i 的这一天接受预约的最大时长
        int[][] dp = new int[len + 1][2];

        // 注意：外层循环从 1 到 =len，相对 dp 数组而言，引用到 nums 数组的时候就要 -1
        for (int i = 1; i <= len; i++) {
            dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1]);
            dp[i][1] = dp[i - 1][0] + nums[i - 1];
        }
        return Math.max(dp[len][0], dp[len][1]);
    }
}

Optimization [Scroll Array] three variables time complexity: complexity of O (N) space: O (1)

public class Solution {

    public int massage(int[] nums) {
        int len = nums.length;
        if (len == 0) {
            return 0;
        }
        if (len == 1) {
            return nums[0];
        }

        // dp[i & 1][0]：区间 [0, i] 里接受预约请求，并且下标为 i 的这一天不接受预约的最大时长
        // dp[i & 1][1]：区间 [0, i] 里接受预约请求，并且下标为 i 的这一天接受预约的最大时长
        int[][] dp = new int[2][2];
        dp[0][0] = 0;
        dp[0][1] = nums[0];

        for (int i = 1; i < len; i++) {
            dp[i & 1][0] = Math.max(dp[(i - 1) & 1][0], dp[(i - 1) & 1][1]);
            dp[i & 1][1] = dp[(i - 1) & 1][0] + nums[i];
        }
        return Math.max(dp[(len - 1) & 1][0], dp[(len - 1) & 1][1]);
    }
}

###### 2. Dynamic Programming a one-dimensional variable-to-back state to think clearly every step of the
time complexity: O (N) space complexity: O (N)

   public int massage(int[] nums) {
        int len = nums.length;
        if (len == 0) {
            return 0;
        }
        if (len == 1) {
            return nums[0];
        }

        // dp[i]：区间 [0, i] 里接受预约请求的最大时长
        int[] dp = new int[len];

      //初始化状态
        dp[0] = nums[0];
        dp[1] = Math.max(nums[0], nums[1]);

        for (int i = 2; i < len; i++) {
            // 今天在选与不选中，选择一个最优的
            dp[i] = Math.max(dp[i - 1], dp[i - 2] + nums[i]);
        }
        return dp[len - 1];
    }

Optimization [Scroll Array] three variables time complexity: complexity of O (N) space: O (1)

class Solution {

    public int massage(int[] nums) {
        int len = nums.length;
        if (len == 0) {
            return 0;
        }
        if (len == 1) {
            return nums[0];
        }

        // dp[i % 3]：区间 [0，i] 里接受预约请求的最大时长
        int[] dp = new int[3];
        dp[0] = nums[0];
        dp[1] = Math.max(nums[0], nums[1]);

        for (int i = 2; i < len; i++) {
            // 今天在选与不选中，选择一个最优的
            dp[i % 3] = Math.max(dp[(i - 1) % 3], dp[(i - 2) % 3] + nums[i]);
        }
        return dp[(len - 1) % 3];
    }
}

【to sum up】

1. Dynamic Planning Process

• Step 1: Design status
• Step 2: state transition equation
• Step 3: Consider initialization
• Step 4: Consider Output
• Step 5: Consider whether compressed state

2. Bottom-up dynamic programming state transition
[General Programming] top-down
[Memory Recursive "at any time may face new problems

Reference link: https: //leetcode-cn.com/problems/the-masseuse-lcci/solution/dong-tai-gui-hua-by-liweiwei1419-8/