1599.经营摩天轮的最大利润

链接：1599.经营摩天轮的最大利润
难度：Medium
标签：数组、模拟
简介：返回最大化利润所需执行的 最小轮转次数 。

题解 1 - cpp

• 编辑时间：2023-03-05
• 执行用时：100ms
• 内存消耗：79.5MB
• 编程语言：cpp
• 解法介绍：模拟。

class Solution {
public:
    int minOperationsMaxProfit(vector<int>& customers, int boardingCost, int runningCost) {
        int resMax = 0, resCnt = -1, wait = 0, cur = 0, i = 0;
        while (wait != 0 || i < customers.size()) {
            if (i < customers.size()) wait += customers[i];
            cur += min(wait, 4) * boardingCost - runningCost;
            wait = max(wait - 4, 0);
            if (cur > resMax) resMax = cur, resCnt = i + 1;
            i += 1;
        }
        return resCnt;
    }
};

题解 2 - python

• 编辑时间：2023-03-05
• 执行用时：1656ms
• 内存消耗：19.1MB
• 编程语言：python
• 解法介绍：同上。

class Solution:
        def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
            resMax, resCnt = 0, -1
            wait, cur, i = 0, 0, 0
            while wait != 0 or i < len(customers):
                if i < len(customers):
                    wait += customers[i]
                cur += min(wait, 4) * boardingCost - runningCost
                wait = max(wait - 4, 0)
                if cur > resMax:
                    resMax = cur
                    resCnt = i + 1
                i += 1
            return resCnt

题解 3 - rust

• 编辑时间：2023-03-05
• 执行用时：24ms
• 内存消耗：2.8MB
• 编程语言：rust
• 解法介绍：同上。

impl Solution {
    pub fn min_operations_max_profit(
        customers: Vec<i32>,
        boarding_cost: i32,
        running_cost: i32,
    ) -> i32 {
        let (mut resMax, mut resCnt, mut wait, mut cur, mut i) = (0, -1, 0, 0, 0);
        while wait != 0 || i < customers.len() {
            if i < customers.len() {
                wait += customers[i];
            }
            cur += wait.min(4) * boarding_cost - running_cost;
            wait = 0.max(wait - 4);
            if cur > resMax {
                resMax = cur;
                resCnt = i as i32 + 1;
            }
            i += 1;
        }
        resCnt
    }
}

题解 4 - python

• 编辑时间：2024-01-01
• 执行用时：1200ms
• 内存消耗：20.89MB
• 编程语言：python
• 解法介绍：模拟。

class Solution:
    def minOperationsMaxProfit(self, customers: List[int], boardingCost: int, runningCost: int) -> int:
        costCnt = costSum = resSum =0
        resCnt = -1
        mask = 0b0000
        wait = 0
        
        def walk():
            nonlocal costCnt, costSum, resSum, resCnt, mask, wait
            cnt = min(wait, 4)
            costSum += cnt * boardingCost - runningCost
            wait -= cnt
            costCnt += 1
            mask = ((mask << 1) | 0b0001) & 0b1111
            if costSum > resSum:
                resCnt, resSum = costCnt, costSum
        for v in customers:
            wait += v
            walk()
        while wait: walk()
        return resCnt