Gas Station

There are n gas stations along a circular route, where the amount of gas at the ith station is gas[i].

You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from the ith station to its next (i + 1)th station. You begin the journey with an empty tank at one of the gas stations.

Given two integer arrays gas and cost, return the starting gas station's index if you can travel around the circuit once in the clockwise direction, otherwise return -1. If there exists a solution, it is guaranteed to be unique

Example 1:

Input: gas = [1,2,3,4,5], cost = [3,4,5,1,2]
Output: 3
Explanation:
Start at station 3 (index 3) and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
Travel to station 4. Your tank = 4 - 1 + 5 = 8
Travel to station 0. Your tank = 8 - 2 + 1 = 7
Travel to station 1. Your tank = 7 - 3 + 2 = 6
Travel to station 2. Your tank = 6 - 4 + 3 = 5
Travel to station 3. The cost is 5. Your gas is just enough to travel back to station 3.
Therefore, return 3 as the starting index.

My Solutions:

可以先遍历一次看看是否有解，如果sum of gas < sum of cost 一定无解。

因为知道一定有解，所以从index=0开始，遍历到最后，每一次如果汽车tank的油量小于0，说明当前的index是不能用的，此时让index+1，清除tank。

也可以优化一下，不需要先遍历一遍判断有没有解，而是多一个在遍历时计算tank total值。最后检查是不是<0。

public int canCompleteCircuit(int[] gas, int[] cost) {
    if (gas == null || cost == null || gas.length == 0 || cost.length == 0) return -1;
    int gasSum = 0, costSum = 0;
    for (int i = 0; i < gas.length; i++) {
        gasSum += gas[i];
        costSum += cost[i];
    }
    if (gasSum < costSum) return -1;

    // 也可以多一个total值 total = 0;
    int tank = 0, index = 0;
    for (int i = 0; i < gas.length; i++) {
        tank += gas[i] - cost[i];
        if (tank < 0) {
            index = i + 1;
            tank = 0;
        }
        // 在这里算总值 total += gas[i] - cost[i];
    }
    // 在这里返回 if (total < 0) return -1;
    return index;
}