leetcode134 Gas Station

思路：

https://leetcode.com/problems/gas-station/discuss/269604/Java-Greedy-thought-process

关键是要想清楚如果从加油站A出发到不了B，那么从A到B之间的任何一个加油站出发也到不了B。

实现：

 #include <bits/stdc++.h>
 using namespace std;
 class Solution
 {
 public:
     int canCompleteCircuit(vector<int>& gas, vector<int>& cost)
     {
         int n = gas.size(), sum = , start = , ans = -;
         for (int i = ; i <  * n; i++)
         {
             sum += gas[i % n];
             sum -= cost[i % n];
             if (sum < ) { sum = ; start = i + ; }
             if (i - start == n) { ans = start; break; }
         }
         return ans;
     }
 };
 int main()
 {
     // int g[] = {1, 2, 3, 4, 5};
     // int c[] = {3, 4, 5, 1, 2};
     // int g[] = {2, 3, 4};
     // int c[] = {3, 4, 3};
     int g[] = {, , , , };
     int c[] = {, , , , };
     vector<int> gas(begin(g), end(g));
     vector<int> cost(begin(c), end(c));
     cout << Solution().canCompleteCircuit(gas, cost) << endl;
     return ;
 }