134. Gas Station

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本文探讨了一辆汽车沿环形路线行驶的问题，该路线上有N个加油站，每个加油站提供一定量的汽油。汽车从其中一个加油站开始，油箱为空，且从一站到下一站的油耗已知。文章提供了两种解决方案来确定汽车能否完成环路行驶，并找到合适的起点。

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.

https://leetcode.com/discuss/16087/space-running-time-solution-anybody-have-posted-this-solution
https://leetcode.com/discuss/4159/share-some-of-my-ideas

这道题的基本上是建立于一个假设之上的：
如果gas[i]的和大于等于 cost[i]的和，那么一定存在一条路径使得从第i个加油站出发，能够走一圈。

方法一：

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        int start = gas.size() - 1;
        int end = 0;
        int sum = gas[start] - cost[start];
        while (start > end) {
            if (sum >= 0) {
                sum += gas[end] - cost[end];
                end++;
            }
            else {
                start--;
                sum += gas[start] - cost[start];
            }
        }
        return sum >= 0 ? start : -1;
    }
};

方法二：

class Solution {
public:
    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
        int start(0), total(0), tank(0);
        //if car fails at 'start', record the next station
        for (int i = 0; i < gas.size(); i++) {
            if ((tank = tank + gas[i] - cost[i]) < 0) {
                start = i + 1;
                total += tank;
                tank = 0;
            }
        }
        return (total + tank < 0) ? -1 : start;
    }
};