[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.
Hide Tags： Greedy
题目：循环一圈的道路上有gasSize个加气站gas[gasSize]，从i到i+1加气站需要耗费cost[i]的气，要求可以从其中任意一个加气站出发，只要能保证回到出发点。
思路：贪恋算法。
　　第一次循环：从gas[0]加气站出发i 从0-> gasSize-1， 挨站判断是否能到第i站，over 为每到一站后剩余的燃气，over 为负数说明骑车到不了当前站i，当发现到不了的情况后当前站，马上就将当前站i作为出发点，over清零，重新开始向前走。
　　第二次循环：当然有可能走完一圈后还不能判断的，因为有可能起点并不是gas[0]，所以还需要继续向前走，直到gas[start ]，期只要有出现到不了的情况这判断为-1。如果能到最后判断over是否大于0，大于0则说明能到，否则返回-1。

int canCompleteCircuit(int* gas, int gasSize, int* cost, int costSize) {
    int i = 0;
    int over = gas[0];
    int start =  0;
    if(gasSize == 1) return (gas[0] >= cost[0]) ?  0 : -1;
    for(i=1;i<gasSize;i++)
    {
        over = over - cost[i-1];
        if(over < 0)
        {
            start = i;
            over = 0;
        }
        over = over + gas[i];
    }

    over = over - cost[gasSize-1];
    for(i=0;i<start;i++)
    {
        if(over < 0)
        {
           return -1;
        }
        over = over + gas[i] - cost[i];
    }
    return (over >= 0) ? start : -1 ; 
}