2023-05-18:有 n 名工人。 给定两个数组 quality 和 wage , 其中,quality[i] 表示第 i 名工人的工作质量,其最低期望工资为 wage[i] 。 现在我们想雇佣

2023-05-18:有 n 名工人。 给定两个数组 quality 和 wage ,

其中,quality[i] 表示第 i 名工人的工作质量,其最低期望工资为 wage[i] 。

现在我们想雇佣 k 名工人组成一个工资组。在雇佣 一组 k 名工人时,

我们必须按照下述规则向他们支付工资:

对工资组中的每名工人,应当按其工作质量与同组其他工人的工作质量的比例来支付工资。

工资组中的每名工人至少应当得到他们的最低期望工资。

给定整数 k ,返回 组成满足上述条件的付费群体所需的最小金额。

输入: quality = [10,20,5], wage = [70,50,30], k = 2。

输出: 105.00000。

答案2023-05-18:

解题步骤:

1.构造 Employee 结构体,存储每个员工的工作质量和垃圾系数(wage / quality)。

2.按照垃圾系数从小到大对所有员工进行排序。

3.维护一个大小为 k 的小根堆,表示当前最低期望工资组中的 k 名工人的工作质量。

4.遍历所有员工,如果堆未满,则将该员工加入堆中并更新最低期望工资。如果堆已满,则检查当前员工能否替换堆顶元素,如果可以,则弹出堆顶元素并将当前员工入堆,更新最低期望工资。

5.最终返回最低期望工资即可。

注意事项:

  • 使用 golang 内置的 container/heap 库来实现小根堆。

  • 在比较垃圾系数大小时,需要使用小于等于号,因为可能存在两个员工的垃圾系数完全相等的情况。

时间复杂度:排序所需的时间为 O(nlogn),遍历员工数组时每个员工会入堆一次,出堆一次,即共进行了 2n 次操作,而小根堆的插入和弹出操作时间复杂度均为 O(logk),因此总时间复杂度为 O(nlogn + nlogk)。

空间复杂度:除给定数组外,我们还需要构造 Employee 结构体,以及维护大小为 k 的小根堆,因此需要额外使用 O(n) 空间来存储结构体数组,并且在堆满时可能需要对堆进行调整,最多需要 O(k) 的额外空间。因此总空间复杂度为 O(n + k)。

go完整代码如下:

package main

import (
	"container/heap"
	"fmt"
	"sort"
)

type Employee struct {
	RubbishDegree float64
	Quality       int
}

func NewEmployee(w, q int) Employee {
	return Employee{RubbishDegree: float64(w) / float64(q), Quality: q}
}

type EmployeeHeap []int

func (h EmployeeHeap) Len() int            { return len(h) }
func (h EmployeeHeap) Less(i, j int) bool  { return h[i] > h[j] }
func (h EmployeeHeap) Swap(i, j int)       { h[i], h[j] = h[j], h[i] }
func (h *EmployeeHeap) Push(x interface{}) { *h = append(*h, x.(int)) }
func (h *EmployeeHeap) Pop() interface{} {
	n := len(*h)
	x := (*h)[n-1]
	*h = (*h)[:n-1]
	return x
}

func min(a, b float64) float64 {
	if a < b {
		return a
	}
	return b
}

func mincostToHireWorkers(quality []int, wage []int, k int) float64 {
	n := len(quality)
	employees := make([]Employee, n)
	for i := range quality {
		employees[i] = NewEmployee(wage[i], quality[i])
	}
	sort.Slice(employees, func(i, j int) bool {
		return employees[i].RubbishDegree <= employees[j].RubbishDegree
	})
	minTops := &EmployeeHeap{}
	heap.Init(minTops)
	ans, qualitySum := 1e9, 0
	for i := 0; i < n; i++ {
		curQuality := employees[i].Quality
		if minTops.Len() < k { // 堆没满
			qualitySum += curQuality
			heap.Push(minTops, curQuality)
			if minTops.Len() == k {
				ans = min(ans, float64(qualitySum)*employees[i].RubbishDegree)
			}
		} else { // 来到当前员工的时候,堆是满的!
			// 当前员工的能力,可以把堆顶干掉,自己进来!
			if top := (*minTops)[0]; top > curQuality {
				heap.Pop(minTops)
				qualitySum += curQuality - top
				heap.Push(minTops, curQuality)
				ans = min(ans, float64(qualitySum)*employees[i].RubbishDegree)
			}
		}
	}
	return ans
}

func main() {
	quality := []int{10, 20, 5}
	wage := []int{70, 50, 30}
	k := 2
	result := mincostToHireWorkers(quality, wage, k)
	fmt.Println(result)
}

在这里插入图片描述

rust完整代码如下:

struct Employee {
    rubbish_degree: f64,
    quality: i32,
}

impl Employee {
    fn new(w: i32, q: i32) -> Self {
        let rubbish_degree = w as f64 / q as f64;
        Self {
            rubbish_degree,
            quality: q,
        }
    }
}

fn mincost_to_hire_workers(quality: Vec<i32>, wage: Vec<i32>, k: i32) -> f64 {
    let n = quality.len();
    let mut employees = Vec::with_capacity(n);
    for i in 0..n {
        employees.push(Employee::new(wage[i], quality[i]));
    }
    // 只根据垃圾指数排序
    // 要价 / 能力
    employees.sort_by(|a, b| a.rubbish_degree.partial_cmp(&b.rubbish_degree).unwrap());
    // 请维持力量最小的前K个力量
    // 大根堆!门槛堆!
    let mut min_tops = std::collections::BinaryHeap::new();
    let mut ans = std::f64::MAX;
    let mut quality_sum = 0;
    for i in 0..n {
        // i : 依次所有员工的下标
        // quality_sum : 进入堆的力量总和!
        // cur_quality当前能力
        let cur_quality = employees[i].quality;
        if min_tops.len() < k as usize {
            // 堆没满
            quality_sum += cur_quality;
            min_tops.push(cur_quality);
            if min_tops.len() == k as usize {
                ans = ans.min(quality_sum as f64 * employees[i].rubbish_degree);
            }
        } else {
            // 来到当前员工的时候,堆是满的!
            // 当前员工的能力,可以把堆顶干掉,自己进来!
            if let Some(top) = min_tops.peek() {
                if *top > cur_quality {
                    quality_sum += cur_quality - min_tops.pop().unwrap();
                    min_tops.push(cur_quality);
                    ans = ans.min(quality_sum as f64 * employees[i].rubbish_degree);
                }
            }
        }
    }
    ans
}

fn main() {
    let quality = vec![10, 20, 5];
    let wage = vec![70, 50, 30];
    let k = 2;
    let result = mincost_to_hire_workers(quality, wage, k);
    println!("{}", result);
}

在这里插入图片描述

c完整代码如下:

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

typedef struct Employee {
    double rubbishDegree;
    int quality;
} Employee;

int cmp(const void* a, const void* b) {
    Employee* pa = (Employee*)a;
    Employee* pb = (Employee*)b;
    if (pa->rubbishDegree < pb->rubbishDegree) {
        return -1;
    }
    else if (pa->rubbishDegree > pb->rubbishDegree) {
        return 1;
    }
    else {
        return 0;
    }
}

double mincostToHireWorkers(int* quality, int qualitySize, int* wage, int wageSize, int k) {
    int n = qualitySize;
    Employee* employees = (Employee*)malloc(n * sizeof(Employee));
    for (int i = 0; i < n; i++) {
        employees[i].quality = quality[i];
        employees[i].rubbishDegree = (double)wage[i] / (double)quality[i];
    }
    qsort(employees, n, sizeof(Employee), cmp);
    int* minTops = (int*)malloc(k * sizeof(int));
    int topIndex = -1;
    double ans = INT_MAX;
    int qualitySum = 0;
    for (int i = 0; i < n; i++) {
        int curQuality = employees[i].quality;
        if (topIndex < k - 1) {
            qualitySum += curQuality;
            minTops[++topIndex] = curQuality;
            if (topIndex == k - 1) {
                ans = qualitySum * employees[i].rubbishDegree;
            }
        }
        else {
            if (minTops[0] > curQuality) {
                qualitySum += curQuality - minTops[0];
                minTops[0] = curQuality;
                ans = qualitySum * employees[i].rubbishDegree;
                // 调整,使堆继续保持最小堆的性质
                for (int j = 0; j < k - 1; j++) {
                    if (minTops[j] > minTops[j + 1]) {
                        int tmp = minTops[j];
                        minTops[j] = minTops[j + 1];
                        minTops[j + 1] = tmp;
                    }
                    else {
                        break;
                    }
                }
            }
        }
    }
    free(employees);
    free(minTops);
    return ans;
}

int main() {
    int quality[] = { 10,20,5 };
    int wage[] = { 70,50,30 };
    int k = 2;
    double result = mincostToHireWorkers(quality, sizeof(quality) / sizeof(int), wage, sizeof(wage) / sizeof(int), k);
    printf("%lf\n", result); 
    return 0;
}

在这里插入图片描述

c++完整代码如下:

#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
#include <limits>

using namespace std;

struct Employee {
    double rubbishDegree;
    int quality;
    Employee() = default;
    Employee(int w, int q) : rubbishDegree(static_cast<double>(w) / static_cast<double>(q)), quality(q) {}
    bool operator<(const Employee& other) const {
        return rubbishDegree < other.rubbishDegree;
    }
};

double mincostToHireWorkers(vector<int>& quality, vector<int>& wage, int k) {
    int n = quality.size();
    vector<Employee> employees(n);
    for (int i = 0; i < n; i++) {
        employees[i] = Employee(wage[i], quality[i]);
    }
    // 只根据垃圾指数排序
    sort(employees.begin(), employees.end());
    // 请维持力量最小的前K个力量
    // 大根堆!门槛堆!
    priority_queue<int> minTops;
    double ans = numeric_limits<double>::max();
    for (int i = 0, qualitySum = 0; i < n; i++) {
        // i : 依次所有员工的下标
        // qualitySum : 进入堆的力量总和!
        // curQuality当前能力
        int curQuality = employees[i].quality;
        if (minTops.size() < k) { // 堆没满
            qualitySum += curQuality;
            minTops.push(curQuality);
            if (minTops.size() == k) {
                ans = min(ans, qualitySum * employees[i].rubbishDegree);
            }
        }
        else { // 来到当前员工的时候,堆是满的!
            // 当前员工的能力,可以把堆顶干掉,自己进来!
            if (minTops.top() > curQuality) {
                //              qualitySum -= minTops.top();
                //              qualitySum += curQuality;
                //              minTops.pop();
                //              minTops.push(curQuality);
                qualitySum += curQuality - minTops.top();
                minTops.pop();
                minTops.push(curQuality);
                ans = min(ans, qualitySum * employees[i].rubbishDegree);
            }
        }
    }
    return ans;
}

int main() {
    vector<int> quality = { 10, 20, 5 };
    vector<int> wage = { 70, 50, 30 };
    int k = 2;
    double result = mincostToHireWorkers(quality, wage, k);
    cout << result << endl; // 105
    return 0;
}

在这里插入图片描述

posted @ 2023-05-18 22:35  福大大架构师每日一题  阅读(20)  评论(0编辑  收藏  举报