牛客网 leetcode 在线编程训练 gas-station

题目：
链接：https://www.nowcoder.com/questionTerminal/3b1abd8ba2e54452b6e18b31780b3635
来源：牛客网

There are N gas stations along a circular route, where the amount of gas at station i isgas[i].
You have a car with an unlimited gas tank and it costscost[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.

思路：
环形天然气站，最多只可能有一个起始站，使站内天然气补给和路途开销能让汽车开一圈。我的思路是，从第0个站开始往后计算，如果到某一站发现车内剩余天然气+站内天然气补给<路途开销，就清零，从下一个起始站再开始计算。
ac后看到讨论区有一个思路非常棒，不会浪费已经计算过的值，如果有兴趣可以戳上方链接直达，题目下方即为讨论区。
下面给出我ac的代码。

代码：class Solution {
public:
int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
int stationNum = gas.size();
int result = 0;
int startStation = -1;
bool flag = true;
for (int i = 0; i < stationNum; i++) {
for (int j = 0; j < stationNum; j++) {
if (i + j < stationNum) {
if (result + gas[i + j] - cost[i + j] >= 0) {
result = result + gas[i + j] - cost[i + j];
} else {
result = 0;
flag = false;
break;
}
} else {
if (result + gas[i + j - stationNum] - cost[i + j - stationNum] >= 0) {
result = result + gas[i + j - stationNum] - cost[i + j - stationNum];
} else {
result = 0;
flag = false;
break;
}
}
}
if (flag)
return i;
else
flag = true;
}
return -1;
}
};