算法题LC17：gas-station

贪心：
题目描述：
环形路上有n个加油站，第i个加油站的汽油量是gas[i].
你有一辆车，车的油箱可以无限装汽油。从加油站i走到下一个加油站（i+1）花费的油量是cost[i]，你从一个加油站出发，刚开始的时候油箱里面没有汽油。
求从哪个加油站出发可以在环形路上走一圈。返回加油站的下标，如果没有答案的话返回-1。
注意：
答案保证唯一。

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.
输入描述：

输出描述：

示例1：
输入：

输出：
        
代码：

//穷举
public class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {
        for(int i=0;i<gas.length;i++){
            if(f(gas,cost,i,0,i)) return i;
        }
        return -1;
    }
    public boolean f(int[] g,int []c,int i,int gas,int start){
        if(gas+g[i]<c[i]) return false;
        int a =(i+1)%g.length;
        if(a==start) return true;
        return f(g,c,a,gas+g[i]-c[i],start);
    }
}