Middle-题目82：134. Gas Station

题目原文：
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.
题目大意：
有n个加油站围成一个圈，其中第i个加油站的油量是gas[i].
你有一个汽车，每从第i个加油站开到第i+1个，要消耗cost[i]的汽油。初始油箱为空，问这个汽车能否开完一圈，如果能则返回一个起点位置，否则返回-1.
题目分析：
首先朴素算法就是从每个节点开始开一圈，看看能不能走到头。这样提交会超时。
那么从discuss看到的一个贪心算法，从节点0开始向前开，如果开完一个站点还有油就向前开，如果油量不足，就把出发站点往回移动一站，并从出发点重新开始开，直到遍历完一圈，如果最后油箱里的油≥0，则返回最后的出发点，否则返回-1.
源码：（language：java）

public class Solution {    public int canCompleteCircuit(int[] gas, int[] cost) {            if (gas == null) {                return -1;            }            // Note: The Solution object is instantiated only once and is reused by each test case.            int count = gas.length;            int n = 0;            int gasInCar = 0;            int begin = 0;            int end = 0;            int i = 0;            while (n < count - 1) {                gasInCar += gas[i] - cost[i];                if (gasInCar >=0) {//forward                    end++;                    i=end;                } else {                    begin--;                    if (begin < 0) {                        begin = count - 1;                    }                    i = begin;                }                n++;            }            gasInCar += gas[i] - cost[i];            if (gasInCar >= 0) {                return begin;            } else {                return -1;            }        }}

成绩：
1ms,beats 5.79%,众数1ms,82.52%
Cmershen的碎碎念：
这个算法的理解和实现都不算难，但暂时没想到如何严格证明其正确性。