code第一部分数组：第二十题：加油站

code第一部分数组：第二十题：加油站

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: �e solution is guaranteed to be unique。

分析

O(N) 的解法是，设置两个变量， sum 判断当前的指针的有效性； total 则判断整个数组是否有
解，有就返回通过 sum 得到的下标，没有则返回 -1。

对于一个循环数组，如果这个数组整体和 SUM >= 0，那么必然可以在数组中找到这么一个元素：从这个数组元素出发，绕数组一圈，能保证累加和一直是出于非负状态。

int gasstation(int gas[],int cost[],int n)
{
    int total=0;
    int j=-1;
    int sum;
    for (int i = 0,sum=0; i < n; ++i)
    {
        sum+=gas[i]-cost[i];
        total+=gas[i]-cost[i];
        if (sum<0)
        {
            j=i;
            sum=0;
        }
    }
    return total>=0?j+1:-1;
}