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.

通过gas - cost 构造一个数组left，left[i]表征 i 站到达下一站之后的剩余gas量；

注意点：1.记得把pos取模，从超出位置拉回来就好； 2. 根据聚合分析，时间复杂度是O(N)，因为在一遍扫描之后就可以确定start位置，之后再追加一个N次扫描，因此至多为2N的时间。代码如下：

class Solution {
public:
    int canCompleteCircuit(vector
  
    &gas, vector
   
     &cost) { int m = gas.size(); int count = 0, sum = 0, pos = 0, start = 0; vector
    
      left; for (int i = 0; i < m; i++) left.push_back(gas[i] - cost[i]); for(int i = 0; i < m; i++) sum += left[i]; if (sum < 0) return -1; sum = 0; while (count < m){ //利用count来记录走过的站点数 sum += left[pos]; pos = ++pos % m; //取模来把超出范围的pos值拉回到0， m % m = 0 既第m+1 个pos又置0 if (sum < 0){ sum = 0; count = 0; start = pos; } else{ count++; } } return start; } };