Gas Station

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本文探讨了一个环形路线油站问题，通过分析每个站点的油量和前往下一个站点的成本，提出了一种算法来确定是否可以从任意一站出发完成整个环形路线。关键在于确保总油量超过总成本，通过遍历每个站点更新剩余油量，最终找到可行的起始点。

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.

1. 如果每个站点的油量总和大于消耗总和，应该总有一个出发点是有效的（无限油箱可以保证这点）；
2. 遍历所有站点，同时更新总油量和总消耗，最后可以知道是否有解
3. 遍历的时候保持一个当前可用油量x，当x小于0，表示从开始点到当前点没有合适的起始点，所以把起始点更新为当前点
4. 最后根据是否有解来返回-1或者最有更新过的x

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class Solution {
public:
int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
int glen = gas.size();
int clen = cost.size();
if (glen != clen) return -1;
int curleft = 0, tleft = 0, rindex = -1;
for (int i = 0; i < glen; i++) {
tleft += gas[i] - cost[i];
tleft += curleft;
if (tleft >= 0 && rindex < 0) {
rindex = i1;
} else if (tleft < 0) {
rindex = -1;
curleft = 0;
}
}
return tleft >= 0 ? rindex : -1;
}
};