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134. Gas Station ​

题目 ​

There are n gas stations along a circular route, where the amount of gas at the ith station is gas[i].

You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from the ith station to its next (i + 1)th station. You begin the journey with an empty tank at one of the gas stations.

Given two integer arrays gas and cost, return the starting gas station's index if you can travel around the circuit once in the clockwise direction, otherwise return -1. If there exists a solution, it is guaranteed to be unique.

 

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.

 

Constraints:

  • n == gas.length == cost.length
  • 1 <= n <= 105
  • 0 <= gas[i], cost[i] <= 104
  • The input is generated such that the answer is unique.
Related Topics
  • 贪心
  • 数组

  • 👍 1860
  • 👎 0

    • 解法一 ​

    暴力枚举解法，每次需要重新计算 tank 的值，不够贪心，超时了

    python
    class Solution:
    def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
        size = len(gas)
        for i in range(size):
            tank = 0
            if gas[i] >= cost[i] and gas[i] > 0:
                tank += gas[i]
                for j in range(size):
                    cur =  i + j  if i + j  < size else i + j - size
                    next = i + j + 1  if i + j  + 1< size else i + j + 1  - size
                    tank = tank  - cost[cur]
                    if tank < 0:
                        break
                    tank +=  gas[next]
                else:
                    return i
        return -1
    • 时间复杂度 O(N2)
    • 空间复杂度 O(1)

    解法二 ​

    只遍历一次，只要 tank 最后大于零即可，可以通过所有的 gas station， 只需要更新首次加油的gas[i] 即可

    py
    from typing import  List
# leetcode submit region begin(Prohibit modification and deletion)

class Solution:
    def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
        total = 0
        tank = 0
        cur = 0

        for i in range(len(gas)):
            diff = gas[i] - cost[i]
            total += diff
            tank += diff
            if tank < 0:
                cur = i + 1
                tank = 0


        return (cur if cur < len(gas) else cur - len(gas) ) if total >=0  else -1
        
# leetcode submit region end(Prohibit modification and deletion)


Solution().canCompleteCircuit([2],	[2]
)

    复杂度分析 ​

    • 时间复杂度 O(N)
    • 空间复杂度 O(1)