文章目錄

• 題目
• 算法思路
• C++代碼

題目

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.

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

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.

算法思路

本周繼續學習動態規劃的思想，本題也是使用動態規劃的案例，不過這個題目尋找恰當的子問題會困難一些。

我們考慮以maxlen(n)作為為經過n個加油站能到達的最遠距離，接下來我們考慮怎么沿著加油站列表更新maxlen(n)的值。

n=0時，顯然maxlen(0) = startfuel

逐步考慮前i個加油站的maxlen(n)數組,當添加一個新的加油站i時，原maxlen數組中maxlen[j] >= stations[i][0]的部分，即能到達新加油站的值能更新后一個值，maxlen[j+1] = （maxlen[j+1]，maxlen[j] + stations[i][1]）

C++代碼

class Solution { public:int minRefuelStops(int target, int startFuel, vector<vector<int>>& stations) {unsigned int maxlen[stations.size()+1] = {0};maxlen[0] = startFuel;for (int i = 0; i < stations.size(); ++i){for (int j = i; j >= 0; --j){if (maxlen[j] >= stations[i][0] && maxlen[j+1] < maxlen[j] + stations[i][1]){maxlen[j+1] = maxlen[j] + stations[i][1];}}}for (int i = 0; i <= stations.size(); ++i){if (maxlen[i] >= target){return i;}}return -1;} };

總結

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