二分优化dp，LeetCode 1235. Maximum Profit in Job Scheduling

一、题目

1、题目描述

2、接口描述

python3

一、题目

1、题目描述

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, return the maximum profit you can take such that there are no two 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.

2、接口描述

python3

复制代码

python 复制代码

class Solution:
    def jobScheduling(self, startTime: List[int], endTime: List[int], profit: List[int]) -> int:

cpp

复制代码

cpp 复制代码

class Solution {
public:
    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {

    }
};

3、原题链接

1235. 规划兼职工作

二、解题报告

1、思路分析

经典区间问题，我们通常处理策略为按照某一端排序

这里按照右端点升序排序

然后定义状态 f[i] 为前 i 个工作所能取得的最大收益

那么f[i + 1] = max(f[i], f[j] + profit[i])

即第 i 个工作选或不选，j要满足endTime[j] <= startTime[i]，这个由于我们已经按照右端点升序排序，所以可以二分查找来快速找到 j

2、复杂度

时间复杂度： O(nlogn)空间复杂度：O(n)

3、代码详解

python3

复制代码

python 复制代码

class Solution:
    def jobScheduling(self, startTime: List[int], endTime: List[int], profit: List[int]) -> int:
        p = sorted(zip(startTime, endTime, profit), key=lambda x:x[1])
        n = len(p)
        f = [0] * (n + 1)
        for i, (s, e, w) in enumerate(p):
            idx = bisect_left(p, s + 1, key=lambda x: x[1], hi = i)
            f[i + 1] = max(f[i], f[idx] + w)
        return f[n]

cpp

复制代码

cpp 复制代码

class Solution {
public:
    int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
        int n = startTime.size();
        vector<array<int, 3>> p(n);
        for (int i = 0; i < n; i ++)
            p[i] = { startTime[i], endTime[i], profit[i] };

        sort(p.begin(), p.end(), [](const auto& a, const auto& b){
            return a[1] < b[1];
        });
        vector<int> f(n + 1);
        for (int i = 0; i < n; i ++){
            int idx = lower_bound(p.begin(), p.begin() + i, array<int, 3>{ 0, p[i][0] + 1, 0}, [](const auto& a, const auto& b){
                return a[1] < b[1];
            }) - p.begin();
            f[i + 1] = max(f[i], f[idx] + p[i][2]);
        }
        return f[n];
    }
};