LeetCode 1235 Maximum Profit in Job Scheduling (dp 排序 二分)

We have n jobs, where every job is scheduled to be done from startTime[i] to endTime[i], obtaining a profit of profit[i].

You're given the startTime , endTime and profit arrays, you need to output the maximum profit you can take such that there are no 2 jobs in the subset with overlapping time range.

If you choose a job that ends at time X you will be able to start another job that starts at time X.

 

Example 1:

Input: startTime = [1,2,3,3], endTime = [3,4,5,6], profit = [50,10,40,70]
Output: 120
Explanation: The subset chosen is the first and fourth job. 
Time range [1-3]+[3-6] , we get profit of 120 = 50 + 70.

Example 2:

Input: startTime = [1,2,3,4,6], endTime = [3,5,10,6,9], profit = [20,20,100,70,60]
Output: 150
Explanation: The subset chosen is the first, fourth and fifth job. 
Profit obtained 150 = 20 + 70 + 60.

Example 3:

Input: startTime = [1,1,1], endTime = [2,3,4], profit = [5,6,4]
Output: 6

 

Constraints:

• 1 <= startTime.length == endTime.length == profit.length <= 5 * 10^4
• 1 <= startTime[i] < endTime[i] <= 10^9
• 1 <= profit[i] <= 10^4

题目链接：https://leetcode.com/problems/maximum-profit-in-job-scheduling/

题目大意：n个任务，每个任务有开始时间，结束时间和收益，同一时刻只能做一个任务，求最大总收益

题目分析：用结构体存任务信息并按结束时间排序，dp[i]表示到第i个任务时的最大值，容易发现存在三种决策方案

1. 不选第i个，则dp[i] = dp[i - 1]

2. 选第i个，已第i个作为起点（时间点与之前的冲突）

3. 选第i个，并接到第j个后面，j为 max(j) task[j].endTime <= task[i].startTime

第三种情况找的时候可以二分

15ms，时间击败82.16%

class Solution {
    
    class Task {
        public int s, e, val;
        public Task(int s, int e, int v) {
            this.s = s;
            this.e = e;
            this.val = v;
        }
    }
    
    public int maxPrePos(int cur, Task[] t) {
        int ans = -1, l = 0, r = cur, mid = 0;
        while (l <= r) {
            mid = (l + r) >> 1;
            if (t[mid].e > t[cur].s) {
                r = mid - 1;
            } else {
                ans = mid;
                l = mid + 1;
            }
        }
        return ans;
    }
    
    public int jobScheduling(int[] startTime, int[] endTime, int[] profit) {
        int n = profit.length;
        Task[] t = new Task[n + 1];
        for (int i = 0; i < n; i++) {
            t[i] = new Task(startTime[i], endTime[i], profit[i]);
        }
        Arrays.sort(t, new Comparator() {
            public int compare(Task a, Task b) {
                if (a == null) {
                    return 1;
                }
                if (b == null) {
                    return -1;
                }
                if (a.e > b.e) {
                    return 1;
                } else if (a.e < b.e) {
                    return -1;
                } else {
                    return a.s > b.s ? 1 : -1;
                }
            }
        });
        // for (int i = 0; i < n; i++) {
        //     System.out.println("s[" + i + "] = " + t[i].s + " e[" + i + "] = " + t[i].e + " v[" + i + "] = " + t[i].val);
        // }
        int[] dp = new int[n + 1];
        dp[0] = t[0].val;
        for (int i = 1; i < n; i++) {
            dp[i] = Math.max(dp[i - 1], t[i].val);
            int pre = maxPrePos(i, t);
            //System.out.println("i = " + i + " pre = " + pre);
            if (pre != -1) {
                dp[i] = Math.max(dp[i], dp[pre] + t[i].val);
            }
            //System.out.println("dp[" + i + "] = " + dp[i]);
        }
        return dp[n - 1];
    }
}