LeetCode _ Gas Station

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.

思路： 类似KMP不回朔的思想。start表示开始的点，sum表示当前汽车的油量。当汽车到达汽油站i时如果不能到达下一站，则更新start直到可以使汽车能够从当前节点到达下一站，如果不存在，则把start设置为下一站。

class Solution {
public:
    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
       int len = gas.size();
       if(cost.size() != len) return -;
       vector<int> flag(len*, );
       for(int i = ; i< len; i++)
            flag[i] = gas[i] - cost[i];
       for(int i = len ; i< len *; ++i)
            flag[i] = flag[i-len];

       int start =  , sum = ;
       for(int i = ; i< len + start && start < len; )
        {
            sum += flag[i] ;
            if(sum >= ){
                ++i;
                continue;
            }
            while(sum<  && start < i){
                sum -= flag[start];
                ++start;
            }
            i++;
            if(sum < ){
                sum = ;
                start = i;
            }

        }
        if(start <len)
         return start;
        return -;

    }
};