- """
- 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.
- """"""
- 解法一: 自己 AC
- 按照题目要求正常实现
- """
- class Solution1:
- def canCompleteCircuit(self, gas, cost):
- cur = 0
- for i in range(len(gas)):
- cur = gas[i]
- for j in range(i, i+len(gas)):
- cur -= cost[j % len(gas)]
- if cur <0:
- break
- cur += gas[(j+1) % len(gas)]
- if cur>= 0:
- return i
- return -1
- """
- 解法二: 算法思想, 主要遵从两个原则:
- 1. 出发点必须满足 gas>cost, 否则不可能出发
- 2. 总的 gas>cost, 否则一定走不完
- """
- class Solution:
- def canCompleteCircuit(self, gas, cost):
- total = 0
- res = 0
- cur = 0
- for i in range(len(gas)):
- cur += gas[i] - cost[i]
- if cur <0:
- res = i + 1
- total += cur
- cur = 0
- return res if (total + cur)>= 0 else -1
- if __name__ == '__main__':
- gas = [1, 2, 3, 4, 5]
- cost = [3, 4, 5, 1, 2]
- ans = Solution1()
- print(ans.canCompleteCircuit(gas, cost))
来源: http://www.bubuko.com/infodetail-3457479.html