LeetCode134. Gas Station

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.

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思路：

题意是有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需要耗费油量为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。

totalGas=gas[0]+…+gas[len];
totalCost=cost[0]+…+cost[len];
如果成功那么totalGas-totalCost>=0。

remain表示从开始点start到当前点i剩下的油量，如果为负，说明start~i这段中没有符合条件的点作为开始点，所以从i+1又开始检查。

代码：

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        if(gas.size()==0 || cost.size()==0){
            return -1;
        }
        int len=gas.size();
        int remain=0;
        int start=0;
        int total=0;
        for(int i=0;i<len;i++){
           total+=(gas[i]-cost[i]);
            if(remain<0){
                remain=gas[i]-cost[i];
                start=i;
            }
            else{
                remain+=(gas[i]-cost[i]);
            }
        }
        return total<0?-1:start;
    }
};