Gas Station

题目：

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.

这个题，一上来就想用穷举法做了。做法是，以每一个点作为起点，都试一次。复杂度O(n^2)。这个。。。不大好。其实我们重复计算了很多。

比如，我们以 a 为起点，试图走 a + 1， a + 2，a + 3 ... 一直走到b。如果我们在计算过程中发现，a 根本到不了b。按照原来的方法，我们该以 a + 1为起点，再算一次看能不能到b。

想想是不可能的。如果从a出发， 到a + 1点后，邮箱里有从a剩下的油（最不济刚好为空）。带着剩下的油都到不了b，你空着邮箱从头开始怎么可能。所以起点肯定不在a到b这段了。直接从b + 1开始算。

public int canCompleteCircuit(int[] gas, int[] cost) {
        if (gas == null || cost == null || gas.length == 0 || cost.length == 0) {
            return -1;
        }
        int start = 0;
        int tank = 0;
        int sum = 0;
        for (int j = 0; j < gas.length; j++) {
            int diff = gas[j] - cost[j];
            if (tank + diff >= 0) {
                tank += diff; 
            }else{
                start = j + 1;
            }
        }
        for (int i = 0; i < gas.length; i++) {
            sum = sum + gas[i] - cost[i];
        }
        if (sum >= 0) {
            return start;
        }else{
            return -1;
        }
    }