134. Gas Station加油站

［抄题］：

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.

 [暴力解法]：

时间分析：

空间分析：

 [优化后]：

时间分析：

空间分析：

［奇葩输出条件］：

［奇葩corner case］：

［思维问题］：

不知道怎么表示：天啦噜，多开几个变量还没学会么？

［英文数据结构或算法，为什么不用别的数据结构或算法］：

［一句话思路］：

tank就一直+=就行了，就可以不必清空 

［输入量］：空： 正常情况：特大：特小：程序里处理到的特殊情况：异常情况（不合法不合理的输入）：

［画图］：

［一刷］：

［二刷］：

［三刷］：

［四刷］：

［五刷］：

  [五分钟肉眼debug的结果]：

［总结］：

tank就一直+=就行了，就可以不必清空。不符合条件的时候才清空。

［复杂度］：Time complexity: O(n) Space complexity: O(1)

［算法思想：迭代/递归/分治/贪心］：

［关键模板化代码］：

［其他解法］：

［Follow Up］：

［LC给出的题目变变变］：

 [代码风格] ：

 [是否头一次写此类driver funcion的代码] ：

 [潜台词] ：

// package whatever; // don't place package name!



class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {
        //corner case
      if (gas == null || cost == null) return -1; 
      
      //initialization
      int sumGas = 0;
      int sumCost = 0;
      int tank = 0;
      int start = 0;
      
      //for loop and renew the start
      for (int i = 0; i < gas.length; i++) {
        sumGas += gas[i];
        sumCost += cost[i];
        tank += gas[i] - cost[i];
        //if tank < 0, renew start
        if (tank < 0) {
          start = i + 1;
          tank = 0;
        }
      }
      
      //if sumgas > sumcost, return start. 
      if (sumGas >= sumCost) return start;
      
      return -1;
    }
}
/*
gas  = [1, 2, 3, 4, 5]
cost = [3, 4, 5, 1, 2]
i      0   1  2  3  4
sumgas 1   3  6  10 15
sumcost 3  7  12 13 15
tank   -2        3  6


*/