【leetcode】Gas Station

题目：

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.

思路：

1：计算出每个站点间gas[i] 和 cost[i] 的差值，然后从正值开始转一圈测试，若差值的结果均为正，则返回起始坐标。但这样做TLE，复杂度O(n^2)；

2：观察上面的过程，很多计算都是重复的。某一段如果能通过的话，那么在所有其他的起点路径中就一定能通过。所以先测试一小段路径，如果能通过的话（也就是gas[i] -cost[i] >= 0）,那么增加这个路径就是测试一下个站点。若果不能通过的话，就改变起点（将起点前移一个站点），使其保证所有gas[i] 和 cost[i]的差值>=0。当测试路径的起点和终点重合时，就能判断是否能够通过了。

代码：

public class Solution {    public int canCompleteCircuit(int[] gas, int[] cost) {        int n = gas.length;        int[] left = new int[n];        int sum,start,current;                for(int i = 0; i < n; i++){        left[i] = gas[i] - cost[i];        }                start = 0;        sum = left[0];        current = 1;        while(current < n && current != start){        if(sum < 0){        start = (start + n - 1)%n;        sum = sum + left[start];        }else{        sum = sum + left[current];        current = (current + 1)%n;        }        }                if(sum >= 0)        return start;        else        return -1;    }}