134. 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 in the clockwise direction, otherwise return -1.

Note:

• If there exists a solution, it is guaranteed to be unique.

• Both input arrays are non-empty and have the same length.

• Each element in the input arrays is a non-negative integer.

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.

Example 2:

Input: 
gas  = [2,3,4]
cost = [3,4,3]

Output: -1

Explanation:
You can't start at station 0 or 1, as there is not enough gas to travel to the next station.
Let's start at station 2 and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
Travel to station 0. Your tank = 4 - 3 + 2 = 3
Travel to station 1. Your tank = 3 - 3 + 3 = 3
You cannot travel back to station 2, as it requires 4 unit of gas but you only have 3.
Therefore, you can't travel around the circuit once no matter where you start.

total代表全部加油站的油減去所有路徑耗費的油，如果全部算完是負值，那說明無解。curSum當前出發點start開始的油量，如果中途變成負值，那麼重設curSum為零，並且更新start的位置為下一個加油站。curSum途中變成零代表從該加油站之前的所有加油站都不可以當作是起點，因為這之間油量變成負數。

// Greedy Algorithm
int canCompleteCircuit(vector<int>& gas, vector<int>& cost) { // time: O(n); space: O(1)
    int total = 0, curSum = 0, start = 0;
    for (int i = 0; i < gas.size(); ++i) {
        total += gas[i] - cost[i];
        curSum += gas[i] - cost[i];
        if (curSum < 0) {
            curSum = 0;
            start = i + 1;
        }
    }
    return total < 0 ? -1 : start;
}