Java C++ 题解leetcode857雇佣K名工人最低成本vector pair
作者:AnjaVon
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题目要求
思路:优先队列 + 贪心
Java
class Solution { public double mincostToHireWorkers(int[] quality, int[] wage, int k) { int n = quality.length; double[][] ratio = new double[n][2]; for (int i = 0; i < n; i++) { ratio[i][0] = wage[i] * 1.0 / quality[i]; ratio[i][1] = quality[i] * 1.0; } Arrays.sort(ratio, (a, b) -> Double.compare(a[0], b[0])); PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> b - a); double res = 1e18; for (int i = 0, tot = 0; i < n; i++) { int cur = (int) ratio[i][1]; tot += cur; pq.add(cur); if (pq.size() > k) tot -= pq.poll(); if (pq.size() == k) res = Math.min(res, tot * ratio[i][0]); } return res; } }
- 时间复杂度:O(n log n)
- 空间复杂度:O(n)
C++
学习了一下vector
和pair
的相互套用,以及自定义排序等内容。
class Solution { public: double mincostToHireWorkers(vector<int>& quality, vector<int>& wage, int k) { int n = quality.size(); vector<pair<double, int>> ratio; for (int i = 0; i < n; i++) { ratio.emplace_back(wage[i] * 1.0 / quality[i], quality[i]); } sort(ratio.begin(), ratio.end(), [](const pair<double, int> &a, const pair<double, int> &b) { return a.first < b.first; }); priority_queue<int> pq; double res = 1e18; for (int i = 0, tot = 0; i < n; i++) { int cur = ratio[i].second; tot += cur; pq.emplace(cur); if (pq.size() > k) { tot -= pq.top(); pq.pop(); } if (pq.size() == k) res = min(res, tot * ratio[i].first); } return res; } };
- 时间复杂度:O(n log n)
- 空间复杂度:O(n)
Rust
use std::collections::BinaryHeap; impl Solution { pub fn mincost_to_hire_workers(quality: Vec<i32>, wage: Vec<i32>, k: i32) -> f64 { let (mut res, mut tot, mut pq) = (f64::MAX, 0, BinaryHeap::new()); let mut ratio = quality.iter().zip(wage.iter()).map(|(q, w)| (*w as f64 / *q as f64, *q as f64)).collect::<Vec<_>>(); ratio.sort_by(|a, b| a.0.partial_cmp(&b.0).unwrap()); for (a, b) in ratio { tot += b as i32; pq.push(b as i32); if pq.len() as i32 > k { tot -= pq.pop().unwrap(); } if pq.len() as i32 == k { res = res.min(a * tot as f64); } } res } }
- 时间复杂度:O(n log n)
- 空间复杂度:O(n)
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