0871. Minimum Number of Refueling Stops - kumaeki/LeetCode GitHub Wiki

0871. 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 <= target, startFuel, stations[i][1] <= 10^9
  • 0 <= stations.length <= 500
  • 0 < stations[0][0] < stations[1][0] < ... < stations[stations.length-1][0] < target

解法1

DP, 依次计算每次加油后能到达的最大距离, 能到达target的最小加油次数就是答案.

class Solution {
    public int minRefuelStops(int target, int startFuel, int[][] s) {
        int len = s.length;
        // dp[i]表示i次加油后到达的最远距离
        long[] pre = new long[len + 1];
        
        pre[0] = startFuel;
        
        // 第i个加油站
        for (int i = 0; i < len; i++){
            long[] cur = new long[len + 1];
            // 第j次加油, 最多加i次油
            for(int j = 0; j <= i; j++){
                cur[j] =  Math.max(pre[j], cur[j]);
                if(s[i][0] > pre[j])
                    continue;
                cur[j + 1] = Math.max(pre[j + 1], pre[j] + s[i][1]);
            }
            pre = cur;
        }
        
        for (int j = 0; j <= len; j++)
            if (pre[j] >= target) 
                return j;
        
        return -1;
    }
}

解法2

Greedy Approach, 每次从可以到达的station中找出加油最多的那一个.

class Solution {
    public int minRefuelStops(int target, int startFuel, int[][] s) {
        Queue<Integer> pq = new PriorityQueue<Integer>((x, y) -> y - x);
        int i = 0, res = 0, distance = startFuel;
        // 当加res次油时
        while (distance < target) {
            
            // 遍历所有的加油站, 对于当前(加res次油)可以到达的加油站
            // 在pq中储存到达加油站后补充的油量
            // 每次循环定义在外层的i都会加1, 保证每次访问的station不会重复
            while (i < s.length && s[i][0] <= distance)
                pq.offer(s[i++][1]);
            
            //pq为空, 所以无法到达任何一个加油站,返回-1
            if (pq.isEmpty()) return -1;
            
            // 在能到达的station中找出加最多油的station, 计算能到达的最远距离
            distance += pq.poll();
            
            res++;
        }
        return res;
    }
}
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