LeetCode:134. Gas Station(Week 7)

134. Gas Station

• 题目
  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.
  

• 解题思路

  • 这道题还是比较简单的，给出汽车行驶到每个点需要的油量cost[i]以及该点可以补充的油量gas[i] ,求能否顺序走完一圈
  • 首先整个行程的总代价total=gas总和 - cost总和，如果total小于0，则整个行程不可能可以被完成，但是如果total大于0，则行程一定可以被完成。
  • preTank记录从开始节点i到当前节点j所得代价
    • preTank < 0，则说明不能顺利到达，因为cost比gas大，当然两个地点中间的任意一点也是不能顺利到达的，因为每次前进的preTank都需要大于等于0，此时需要把当前节点作为行程的节点，并且preTank = gas[j] - cost[j]。
    • preTank >= 0， preTank += gas[j] - cost[j]，继续前进。

• 实现代码

  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;
      }
  };