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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.

来源：力扣（LeetCode）

链接：https://leetcode-cn.com/problems/gas-station 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

gas数组表示加油站的油，cost数组表示此加油站到下一个加油站的耗油，问从哪里出发，可以绕整个路程一周。

有count来表示所经过加油站的剩下的油，count += gas[i] - cost[i];

用tmpCount来统计，从上一个可以加油的加油站，即gas[i] - cost[i] >0的加油站，开始计算经过每个加油站剩下的油，如果tmpCount < 0，则重新计算。

class Solution {    public int canCompleteCircuit(int[] gas, int[] cost) {        int count = 0;        int index = -1;        int tmpCount = -1;        int i = 0;        while (i < gas.length) {            int diff = gas[i] - cost[i];            count += diff;            tmpCount += diff;            if (diff >= 0 && index == -1) {                index = i;                tmpCount = diff;            }            // 表示从index开始的加油站不能满足经过当前加油站i，则重新计算。            if (tmpCount < 0) {                index = -1;            }            i++;        }        if (count < 0) {            return -1;        }        return index;    }}