134. Gas Station

• 題目
  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.
  

• 解題思路

  • 這道題還是比較簡單的，給出汽車行駛到每個點需要的油量cost[i]以及該點可以補充的油量gas[i] ,求能否順序走完一圈
  • 首先整個行程的總代價total=gas總和 - cost總和，如果total小於0，則整個行程不可能可以被完成，但是如果total大於0，則行程一定可以被完成。
  • preTank記錄從開始節點i到當前節點j所得代價
    • preTank < 0，則說明不能順利到達，因為cost比gas大，當然兩個地點中間的任意一點也是不能順利到達的，因為每次前進的preTank都需要大於等於0，此時需要把當前節點作為行程的節點，並且preTank = gas[j] - cost[j]。
    • preTank >= 0， preTank += gas[j] - cost[j]，繼續前進。

• 實現程式碼

  class Solution {
  public:
      int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
          int size = gas.size();
          int preTank = 0;
          int tank = 0;
          int start = 0;
          for(int i = 0; i < size; ++i) {
          	tank += gas[i] - cost[i];
          	if(preTank < 0) {
          		start = i;
          		preTank = gas[i] - cost[i];
          	}
          	else {
          		preTank += gas[i] - cost[i];
          	}
          }
          return tank >= 0 ? start : -1;
      }
  };