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.
class Solution: def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int: total = 0 summ = 0 start = 0 for i in range(len(gas)): total += gas[i]-cost[i] summ += gas[i]-cost[i] if summ<0: summ = 0 start = i+1 continue if total>=0: return start return -1
虽然题目中说的是一个环，但是其实可以不必考虑环，直接就是list就行，这里的total是到达第(i+1)个加油站时的存油量，初始化为0,summ 是到达下一个站加油站时的存油量，如果它为负的话就把它清0,因为这时候它不了下一个加油站，需要重设起点，并把起点更新为下一个加油站. 最终total如果非负的话，说明就可以走完，否则的话就走不完。