题目描述

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

思路一：动态规划。对每个站点i，遍历它可以作为所有可能的第j个站点的情况。j从1到i+1。dp[i]表示用i个站点，能到达的最远距离。那么，dp[j] = dp[j-1]+stop[i].fuel。前提是，dp[j-1] >= stop[i].pos。因为每个站点只能用一次，注意j要从大到小遍历。
思路二：每次油料不够时，其实都是从之前的所有站点中，取油量最大的来加，所以用优先队列存储当前位置可以加的所有油量。每次不够时，取最大油量。

代码

动态规划：

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

优先队列：

class Solution {public:int minRefuelStops(int target, int startFuel, vector<vector<int>>& stations) {int cur = startFuel;int stop = 0;priority_queue<int, vector<int>, less<int> > que;int i = 0;while(true) {while(i < stations.size() && stations[i][0] <= cur) {que.push(stations[i++][1]);}if (cur >= target) return stop;if (que.empty()) break;cur += que.top();que.pop();stop++;}return -1;}
};

自己想的时候一点思路都没有。关键是状态没想到。也没理解好，油量都是累加的，只需要加最多的油量就好了。

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【LeetCode 871】 Minimum Number of Refueling Stops

题目描述

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