Gas Station

Gas Station

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

You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from stationi 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.

public int canCompleteCircuit(int[] gas, int[] cost) {        int loc[] = new int[gas.length];        int i = 0, j = 0, rem = 0, count=0;        while (count < gas.length && j<gas.length) {            if (loc[j] == gas.length) return j;            if (rem + gas[i] >= cost[i]) {                rem += (gas[i] - cost[i]);                loc[j]++;            } else {                rem = 0;                j = i+1;                count++;            }            i=(i+1)%gas.length;        }        return -1;    }

考虑1号加油站，直接模拟判断它是否为解。如果是，直接输出；如果不是，说明在模拟的过程中遇到了某个加油站p，在从它开到加油站p+1时油没了。这样，以2, 3,…, p为起点也一定不是解。这样，使用简单的枚举法便解决了问题，时间复杂度为O(n)