[LeetCode] 134. Gas Station

O(n^2) 的复杂度好像有些高，不知道有没有更快的方法解决 \(▔▽▔)/

	class Solution {
	public int canCompleteCircuit(int[] gas, int[] cost) {
	for (int i = 0; i < gas.length; i++) {
	int gasInCar = gas[i];
	for (int traveled = 1; traveled <= gas.length; traveled++) { // Attn: Made mistake here. traveled should be range from 1 to gas.length (rather than 1 to gas.length - 1)
	int j = (i + traveled) % gas.length;
	int currentCost = j == 0 ? cost[cost.length - 1] : cost[j - 1];
	gasInCar = gasInCar - currentCost;
	if (gasInCar < 0) {
	break;
	}
	gasInCar = gasInCar + gas[j];
	}
	if (gasInCar >= 0) {
	return i;
	}
	}
	return -1;
	}
	}

view raw canCompleteCircuit.java hosted with ❤ by GitHub

更新：找到了很巧妙的方法。确实一次遍历就可以解决。～(￣▽￣～)(～￣▽￣)～
https://blog.csdn.net/JackZhang_123/article/details/78008439