[LeetCode] 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, otherwise return -1.

Note:

The solution is guaranteed to be unique.
使用贪心算法。规律是，如果从某个加油站出发，如果可以到达的最远终点不是自身，那么从这个加油站到其最远终点之间的所有加油站都可以排除，直接以上一轮的终点为起点再开始考虑。因为如果以中途加油站为起点考虑，那样会损失从之前加油站来的时候余下的油的优势，就更不可能最后到达自身了。

如果这样一直考虑，最后终点过了最原始起点，说明所有的点都被考虑过了一遍。时间复杂度O(N)。

public int canCompleteCircuit(int[] gas, int[] cost) {
int length = gas.length;

int start;
int k = 0; // start testing from the first station
while (true) {
int sum = 0;
start = k % length; // save the starting point
do {
sum += gas[k % length] - cost[k % length];
k++;
} while (sum >= 0 && k < start + length);

// complete a cycle
if (k == start + length && sum >= 0) {
return k - length;
}
// test from the end that has not been reached
else {
if (k >= length)
return -1;
}
}
}