leetcode Gas Station

Gas Station

题目：

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.

解析：

1 计算空车从某个station出发，到达下一个station的gasLeft。

2 以某个station为起点进行遍历。

3 计算从起点到达每个station的gasLeft，如果有负值，则不符合条件。

4 想不到竟然Accept了。

代码如下：

    private boolean canCompleteCircuit(int[] gasLeft, int startStation) {
        int gas = 0;
        int currentStation = startStation;
        while(true) {
            gas += gasLeft[currentStation];
            if (gas < 0) {
                return false;
            }
            currentStation ++;
            if (currentStation == gasLeft.length) { currentStation = 0; }
            if (currentStation == startStation) { break; }
        }
        return true;
    }
    public int canCompleteCircuit(int[] gas, int[] cost) {
        int[] gasLeft = new int[gas.length];
        for (int i = 0; i < cost.length; i++) {
            gasLeft[i] = gas[i] - cost[i];
        }
        for (int i = 0; i < gasLeft.length; i++) {
            if (gasLeft[i] >= 0 && canCompleteCircuit(gasLeft, i)) {
                return i;
            }
        }
        return -1;
    }

原本以为会超时。

其实最开始想到的是，汽车可以选择方向。先判断可能travel around的方向，再进行遍历。

这题看起来会有更好解法，请留言。