[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. 每一站的代價爲gas-cost, 也就是求從哪一站開始累加代價和總是大於0。一開始用了一個O（n^2）的解法，超時

2. 如果所有站的代價和大於0，則所求的路線必定存在。如果總代價〉=0,從序號0開始求代價和，如果代價和小於0，則不是從本站或者本站之前的某一個代價大於0的站開始，必從下一站即之後的站開始，而且這樣的站必定存在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;
    }

};