leetcode 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.

给出一个空tank，n个gas station，每经过一站可以加到gas[i]的油，到下一站需要消费掉cost[i]的油，问是否存在一站，从这里出发还可以回到这一站

思路：
最容易想到的就是一站一站地去试，从i出发，走不回i的话就再从i+1试，这种方法会time limit exceed

然后就要改变思维，用greedy
用diff[i]来表示gas[i] - cost[i]

所以每经过一站，tank+=diff[i]
如果从i站走到k-1站的时候tank都是>0的，但是到第k站tank<0了
那么还需要从i+1开始试吗？不需要
因为到k站时tank<0，从i开始到k相当于
diff[i] + diff[i+1] +…+ diff[k]=sum < 0
如果从i+1开始，相当于
diff[i+1] + … + diff[k] = sum - diff[i] < 0

所以一直到k站都不需要再试了，直接从k+1开始

下一个问题，从k+1开始，到数组右边界的时候是不是还需要回到index 0然后再加到k+1？
答案也是不需要，因为刚开始从0出发到k+1的时候，0到k+1部分的和已经计算过了，走完全程其实就相当于计算整个数组元素的和，只要整个数组元素的和>=0，就说明是有可能走完全程的

代码：

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