Minimum Number of Refueling Stops

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. 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

题目理解：

汽车每走一公里耗油一升，初始车里有油，途中会有加油站，加油站里有一定数量的油，问最少在几个加油站停留加油可以到达目的地

解题思路：

使用动态规划来解题。dp[i][j]表示经过i个加油站，使用的最后一个加油站是前j个（含）加油站中的某一个的时候，行驶的最远距离是多少。有了这个定义的话，那么经过i + 1个加油站，而且最后一个加油站是k的情况就可以很容易的推出来，也就是，如果经过i个加油站，而且最后一个加油站在k之前的时候，最远行驶距离可以到达第k个加油站，那么dp[i + 1][k] = dp[i][k - 1] + stations[k][1]。就有了递推公式：

dp[i][j] = dp[i - 1][j - 1] + stations[j][1]        if dp[i - 1][j - 1] > stations[j][0]

dp[i][j] = max{dp[i]j[j - 1], dp[i][j]}

代码如下：

class Solution {
    public int minRefuelStops(int target, int startFuel, int[][] stations) {
        int num_station = stations.length;
        int[][] record = new int[num_station + 1][num_station + 1];
        Arrays.fill(record[0], startFuel);
        if(startFuel >= target)
        	return 0;
        for(int i = 1; i < num_station + 1; i++) {
        	for(int j = 1; j < num_station + 1; j++) {
        		if(record[i - 1][j - 1] >= stations[j - 1][0])
        			record[i][j] = record[i - 1][j - 1] + stations[j - 1][1];
        		if(record[i][j] >= target)
        			return i;
        		record[i][j] = Math.max(record[i][j], record[i][j - 1]);
        	}
        }
        return -1;
    }
}