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. 1 <= target, startFuel, stations[i][1] <= 10^9
2. 0 <= stations.length <= 500
3. 0 < stations[0][0] < stations[1][0] < ... < stations[stations.length-1][0] < target

题目大意：

邮箱容量无限的小车从X轴原点出发向target行驶，初始油量为startFuel，每行驶1个距离单位耗油1L。

途中有若干加油站，记为stations，由二元组(loc, fuel)表示，分别为加油站的位置和油量。

小车每经停一个加油站就可以获得加油站的所有油料。

求小车到达目标位置最少经停多少个加油站，若无法抵达，则返回-1。

解题思路：

解法I 动态规划（Dynamic Programming）

时间复杂度 O(N^2) N为加油站的个数

dp[i][j]表示到达第i个加油站，经停j次时的最大油量

状态转移方程：

dp[i][j] = max(dp[i][j], dp[i - 1][j] - loc[i] + loc[i - 1])

dp[i][j + 1] = max(dp[i][j + 1], dp[i - 1][j] - loc[i] + loc[i - 1] + fuel[i])

Python代码：

class Solution(object):
    def minRefuelStops(self, target, startFuel, stations):
        """
        :type target: int
        :type startFuel: int
        :type stations: List[List[int]]
        :rtype: int
        """
        INF = 10**20
        stations.append([target, 0])
        dp = {0 : startFuel}
        last = 0
        for idx, (loc, fuel) in enumerate(stations):
            ndp = {}
            for dstop, dfuel in dp.items():
                dloc = loc - last
                nfuel = dfuel - dloc
                if dfuel >= dloc:
                    ndp[dstop] = max(ndp.get(dstop, 0), nfuel)
                    ndp[dstop + 1] = max(ndp.get(dstop + 1, 0), nfuel + fuel)
            last = loc
            dp = ndp
        return min(dp.keys() or [-1])

解法II 优先队列（Priority Queue）

时间复杂度 O(N * log N)  N为加油站的个数

优先队列pq（大顶堆）维护当前油量可以抵达的所有加油站。

油量不足以到达下一个加油站时，从pq中弹出堆顶，并令计数器+1

Java代码：

class Solution {
    public int minRefuelStops(int target, int startFuel, int[][] stations) {
        PriorityQueue<Integer> pq = new PriorityQueue<>(Collections.reverseOrder());
        int loc = 0, stop = 0, fuel = startFuel;
        for (int[] st : stations) {
            if (fuel + loc >= target) {
                return stop;
            }
            int sloc = st[0], sfuel = st[1];
            while (!pq.isEmpty() && fuel < sloc - loc) {
                fuel += pq.poll();
                stop += 1;
            }
            if (fuel < sloc - loc) {
                return -1;
            }
            fuel -= sloc - loc;
            pq.add(sfuel);
            loc = sloc;
        }
        while (!pq.isEmpty() && fuel < target - loc) {
            fuel += -pq.poll();
            stop += 1;
        }
        return fuel < target - loc ? -1 : stop;
    }
}