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 in the clockwise direction, otherwise return -1.

Note:

• If there exists a solution, it is guaranteed to be unique.
• Both input arrays are non-empty and have the same length.
• Each element in the input arrays is a non-negative integer.

Example 1:

Input: 
gas  = [1,2,3,4,5]
cost = [3,4,5,1,2]

Output: 3

Explanation:
Start at station 3 (index 3) and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
Travel to station 4. Your tank = 4 - 1 + 5 = 8
Travel to station 0. Your tank = 8 - 2 + 1 = 7
Travel to station 1. Your tank = 7 - 3 + 2 = 6
Travel to station 2. Your tank = 6 - 4 + 3 = 5
Travel to station 3. The cost is 5. Your gas is just enough to travel back to station 3.
Therefore, return 3 as the starting index.

Example 2:

Input: 
gas  = [2,3,4]
cost = [3,4,3]

Output: -1

Explanation:
You can't start at station 0 or 1, as there is not enough gas to travel to the next station.
Let's start at station 2 and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
Travel to station 0. Your tank = 4 - 3 + 2 = 3
Travel to station 1. Your tank = 3 - 3 + 3 = 3
You cannot travel back to station 2, as it requires 4 unit of gas but you only have 3.
Therefore, you can't travel around the circuit once no matter where you start.

题目大意：

一个环路有n个加油站，gas给出每一站能加多少油，cost给出该站到下一站会花费多少油。

解法：

使用tmp存储每一站gas-cost的值，可知当tmp[i]小于0的时候，车不能从这一站出发，因为油不够出发的。

1.首先判断所有tmp的和是否是大于等于0的，如果是，则说明存在解，否则不存在解。

2.再寻找出发的位置，如果说当前位置的tmp[start]<0,则不能从start出发，要找到满足tmp[start]>=0的start

3.sum记录当前剩余油的总量，如果说sum<0，那么start要从end的下一个出发

java:

class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {
        int []tmp=new int[gas.length];
        for(int i=0;i<gas.length;i++){
            tmp[i]=gas[i]-cost[i];
        }
        int sum=0;
        for(int i=0;i<gas.length;i++) sum+=tmp[i];
        if(sum<0) return -1;
        sum=0;
        int start=0,end=0;
        while(start<gas.length){
            if(tmp[start]<0) {start++;end=start;}
            sum+=tmp[end];
            if((end+1)%gas.length==start) break;
            if(sum<0) {
                start=end+1;
                sum=0;
            }
            end=(end+1)%gas.length;
        }

        return start;
    }
}