[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：

1. 每一站的代价为gas-cost, 也就是求从哪一站开始累加代价和总是大于0。一开始用了一个O（n^2）的解法，超时

2. 如果所有站的代价和大于0，则所求的路线必定存在。如果总代价〉=0,从序号0开始求代价和，如果代价和小于0，则不是从本站或者本站之前的某一个代价大于0的站开始，必从下一站即之后的站开始，而且这样的站必定存在O（n）

class Solution {public:    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {        // Note: The Solution object is instantiated only once and is reused by each test case.        int sum=0;        int total=0;        int start=0;        for(int i=0;i<gas.size();i++)        {            sum+=gas[i]-cost[i];            total+=gas[i]-cost[i];            if(sum < 0)            {                start=(i+1)%gas.size(); sum=0;            }        }                if(total <0)            return -1;        else            return start;    }};