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.

分析

题目描述：有N个加油站点构成一个环状，每个站点i加油量为gas[i]，从站点i到站点i+1需要耗费有两为cost[i]，现要求从哪个站点出发可以成功转一圈回到初始站点，返回该站点，若没有则返回-1；

详细思路分析参考：算法分析网址

AC代码

class Solution {

public:

    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {

        if (gas.size() == 0 || cost.size() == 0 || gas.size() != cost.size()) return -1;

        int total = 0, sum = 0, start = 0;

        for (int i = 0; i < gas.size(); ++i){

            total += (gas[i] - cost[i]);

            if (sum < 0){ //发现油箱空了，从下一个站点尝试

                sum = (gas[i] - cost[i]);

                start = i;

            }

            else

                sum += (gas[i] - cost[i]);

        }

        return total < 0 ? -1 : start; //用total判断start 是否是满足要求的解

    }

};