Day 50, ##, Gas Station

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.
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O(n) time, in space
if gas[i] < cost[i], skip to i + 1, reset everything
else, add the difference to the gas[i + 1] station, repeat
until all stations have been checked.
 Note that count < size + 1 since this is supposed to be a loop

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 startPoint = -1, gasLeft = 0;
        int size = gas.size();
        for (int i = 0, count = 0; i < size && count < size + 1; i++, count++) {
            if (i == startPoint) return i; // found a start point
            gasLeft += gas[i];
            if (gasLeft >= cost[i]) {
                if (startPoint == -1) {
                    startPoint = i;
                }
                gasLeft -= cost[i];
                if (i == size - 1) {
                    i = -1;
                }
            }else { // reset 
                gasLeft = 0;
                startPoint = -1;
            }
        }
        return startPoint;
    }
};

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Neat version, from internet

int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
  int sum = 0;
  int total = 0;
  int j = -1;
  for(int i = 0; i < gas.size() ; ++i){
    sum += gas[i]-cost[i];
    total += gas[i]-cost[i];
    if(sum < 0){
      j=i; sum = 0; 
    }
  }
  return total>=0? j+1 : -1;
}

Update on Oct-29-2014
If car starts at A and can not reach B. Any station between A and B can not reach B.(B is the first station that A can not reach.
The solution below is not completely correct, it cannot pass the case where total cost is larger than gas, for instance: gas = {5,5,1,2,4} cost = {4,4,4,4,4}. Although OJ accepts it

public:
    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
        int leftGas = 0;
        int index = -1;
        
        for (int i = 0; i < gas.size(); i++) {
            if (index == -1) index = i;
            leftGas += gas[i] - cost[i];
            if (leftGas < 0) {
                index = -1;
                leftGas = 0;
            }
        }
        
        if (leftGas + gas[0] - cost[0] < 0) return -1;
        
        return index;
    }
};