Gas Station

出自

http://blog.youkuaiyun.com/lbyxiafei/article/details/12183461

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.

首先想到的方法是O(n^2)的循环检测方法，移动过程的位置需要取余操作，(++curPos)%gas.length

这种方法有一个大的test case无法通过。

O(n)解法：

建立两个变量sum和total，从0到length遍历gas[]和cost[]，sum和total都等于gas[i]-cost[i]，检测sum是否小于0，如果是，res变量等于当前index，同时sum归零。

最终根据total变量是否大于零决定返回res或者-1。

很牛逼的做法。

	public int canCompleteCircuit2(int[] gas, int[] cost) {
		// Note: The Solution object is instantiated only once and is reused by
		// each test case.
		int sum = 0, total = 0, len = gas.length, index = -1;
		for (int i = 0; i < len; i++) {
			sum += gas[i] - cost[i];
			total += gas[i] - cost[i];
			if (sum < 0) {
				index = i;
				sum = 0;
			}
		}
		return total >= 0 ? index + 1 : -1;
	}

sum判断当前的指针的有效性；total则判断了整个array是否有解，有就返回通过sum得到的解，没有返回-1。