題目描述

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：動態規劃。對每個站點i，遍歷它可以作為所有可能的第j個站點的情況。j從1到i+1。dp[i]表示用i個站點，能到達的最遠距離。那么，dp[j] = dp[j-1]+stop[i].fuel。前提是，dp[j-1] >= stop[i].pos。因為每個站點只能用一次，注意j要從大到小遍歷。
思路二：每次油料不夠時，其實都是從之前的所有站點中，取油量最大的來加，所以用優先隊列存儲當前位置可以加的所有油量。每次不夠時，取最大油量。

代碼

動態規劃：

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;} };

優先隊列：

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;} };

自己想的時候一點思路都沒有。關鍵是狀態沒想到。也沒理解好，油量都是累加的，只需要加最多的油量就好了。

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