LeetCode-Gas Station(加油站问题)-蒲公英云

LeetCode-Gas Station(加油站问题)

青旅半醒 2022-08-06 16:19 267阅读 0赞

There are N gas stations along a circular route, where the amount of gas at station i is gas[i].

You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to its next station (i+1). You begin the journey with an empty tank at one of the gas stations.

Return the starting gas station’s index if you can travel around the circuit once, otherwise return -1.

Note:

The solution is guaranteed to be unique.

第一想法肯定是暴力破解了，两个循环就搞定了：

public int canCompleteCircuit1(int[] gas, int[] cost) {
        for (int i = 0; i < gas.length; i++) {
            int sum = 0;
            for (int j = i; j < i+gas.length; j++) {
                int station = j%gas.length;
                sum += gas[station] - cost[station];
                if (sum < 0) {
                    break;
                }
            }
            if (sum >= 0) {
                return i;
            }
        }
        return -1;
    }

外层循环是第i加油站，内层循环计算是否能跑完一圈。结果直接超时，需要做优化，根据题目，我们很容易知道：

1）如果gas[]的总和大于cost[]的总和，这个题目肯定有解

2）在1）的前提下，如果有A—>B—>C中A不能到达C，那么B也不能到达C:

证明：已知：gas(A) >= cost(A) (如果gas(A) < cost(A)，则把A当做起点肯定失败，因为他连B都到达不了)

如果：gas(A)+gas(B) < cost(A)+cost(B)

则证：gas(B) < cost(B)

根据2），我们每次发现从A出发不能到达C时，不用把B当做起点计算。只需要从C当做起点继续计算。代码如下：

public int canCompleteCircuit(int[] gas, int[] cost) {
        int total = 0, tank = 0, index = 0;
        for (int i = 0; i < gas.length; i++) {
            int temp = gas[i] - cost[i];
            tank += temp;
            if (tank < 0) {
                tank = 0;
                index = i+1;
            }
            total += temp;
        }
        return total < 0 ? -1 : index;
    }