原题 题意： 过一个循环的加油站，每个加油站可以加一定数量的油，走到下一个加油站需要消耗一定数量的油，判断能否走一圈。
思路： 一开始思路就是遍历一圈，最直接的思路。
class Solution { public: int canCompleteCircuit(vector<int> &gas, vector<int> &cost) { int beg = 0; int tank = 0; int n = gas.size(); int sumGas = 0, sumCost = 0; for (auto a : gas) sumGas += a; for (auto a : cost) sumCost += a; if (sumGas < sumCost) { return -1; } for (int i = 0; i < n; i++) { tank += gas[i]; if (tank >= cost[i]) { tank -= cost[i]; } else { beg = i + 1; tank = 0; } } return beg; } };看到别人的题解，很简洁。
class Solution { public: int canCompleteCircuit(vector<int> &gas, vector<int> &cost) { int max = 0; int sum = 0; int beg = 0; for (int i = gas.size() - 1; i >= 0; i++) { sum += gas[i] - cost[i]; if (max < sum) { max = sum; beg = i; } } return sum < 0 ? -1 : beg; } };转载于:https://www.cnblogs.com/ruoh3kou/p/9893426.html
