LeetCode 871. Minimum Number of Refueling Stops 动态规划类似背包贪心

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这篇博客探讨了一个关于汽车行驶途中加油的问题，通过两种不同的算法思路进行解决。第一种方法利用动态规划，计算每到达一个加油站时的最大行驶距离；第二种方法采用最大堆，保存历史最大油量，以便在油量不足时进行补给。博客通过具体的例子解释了这两种方法的实现，并展示了如何找到最少的加油次数以达到目标位置。

A car travels from a starting position to a destination which is target miles east of the starting position.

Along the way, there are gas stations. Each station[i] represents a gas station that is station[i][0] miles east of the starting position, and has station[i][1] liters of gas.

The car starts with an infinite tank of gas, which initially has startFuel liters of fuel in it. It uses 1 liter of gas per 1 mile that it drives.

When the car reaches a gas station, it may stop and refuel, transferring all the gas from the station into the car.

What is the least number of refueling stops the car must make in order to reach its destination? If it cannot reach the destination, return -1.

Note that if the car reaches a gas station with 0 fuel left, the car can still refuel there. If the car reaches the destination with 0 fuel left, it is still considered to have arrived.

Example 1:

Input: target = 1, startFuel = 1, stations = []
Output: 0
Explanation: We can reach the target without refueling.

Example 2:

Input: target = 100, startFuel = 1, stations = [[10,100]]
Output: -1
Explanation: We can't reach the target (or even the first gas station).

Example 3:

Input: target = 100, startFuel = 10, stations = [[10,60],[20,30],[30,30],[60,40]]
Output: 2
Explanation: 
We start with 10 liters of fuel.
We drive to position 10, expending 10 liters of fuel.  We refuel from 0 liters to 60 liters of gas.
Then, we drive from position 10 to position 60 (expending 50 liters of fuel),
and refuel from 10 liters to 50 liters of gas.  We then drive to and reach the target.
We made 2 refueling stops along the way, so we return 2.

Note:

1 <= target, startFuel, stations[i][1] <= 10^9
0 <= stations.length <= 500
0 < stations[0][0] < stations[1][0] < ... < stations[stations.length-1][0] < target

------------------------------------

思路一：

利用f[i]表示加i次油最远可以走多远，那么当遇到第i个加油站时，需要由f[0],f[1]...f[i-1]来刷新新的f[0],f[1]...f[i]，当然要逆着刷，非常类似背包。而且要求所有加油站都看一遍，最终挑f[i]>=target的最小i

class Solution:
    def minRefuelStops(self, target: int, startFuel: int, stations: List[List[int]]) -> int:
        l = len(stations)
        f = [0 for i in range(l + 1)]
        f[0] = startFuel
        for i in range(l):  # arrive at station i, at most i+1 refueling
            if (f[i] < stations[i][0]):
                return -1
            for j in range(i, -1, -1):
                if (f[j] >= stations[i][0]):
                    f[j + 1] = max(f[j + 1], f[j] + stations[i][1])
                else:
                    break
        for i in range(l+1): #bug2: put this cycle inside
            if (f[i] >= target):
                return i
        return -1

思路二：如果到达i时的油量够，那么就先不加了，放到最大堆里，油不够的时候再从历史最大的里挑：

import heapq
class Solution:
    def minRefuelStops(self, target: int, startFuel: int, stations: List[List[int]]) -> int:
        stations.append([target,0])
        l,cur,history,res = len(stations),startFuel,[],0
        for i in range(l):
            while (history and cur<stations[i][0]):
                cur -= heapq.heappop(history)
                res += 1
            if (cur >= target):
                return res
            if ((not history) and cur<stations[i][0]): #bug1: if (not history)
                return -1
            else:
                heapq.heappush(history, -stations[i][1])
        return res