想到了提前判断和小于0的情况，懒得写，果然被阴间用例10万个加油站坑了。
class Solution:def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:#1n = len(gas)if n ==1:if gas[0] >= cost[0]:return 0else:return -1#-1startpoint =[gas[x] - cost[x] for x in range(n)]if all(element < 0 for element in startpoint) or sum(startpoint) <0:return -1#elseret = -1for i, element in enumerate(startpoint):if element >= 0:total = 0temp =startpoint[i:]+ startpoint[:i]for i2, element2 in enumerate(temp):total += element2if total <0:breakif i2 == n-1:ret = ireturn retreturn ret
然后发现更阴间的99999个0的测试用例,Pycharm调试都崩溃了。。。
跳过了0的情况，题目说明可能的解是唯一的，所以0不影响结果。
class Solution:def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:#1n = len(gas)if n ==1:if gas[0] >= cost[0]:return 0else:return -1#-1startpoint =[gas[x] - cost[x] for x in range(n)]if all(element < 0 for element in startpoint) or sum(startpoint) <0:return -1#elseret = -1for i, element in enumerate(startpoint):if element > 0:total = 0temp =startpoint[i:]+ startpoint[:i]for i2, element2 in enumerate(temp):if element2 ==0:continuetotal += element2if total <0:breakif i2 == n-1:ret = ireturn retreturn ret
然后就遇到99995个1。。。换个思路解题。
不需要考虑已经走过的加油站，只需要维护一个总的燃油量，如果循环结束总量大于0，就说明已经走过的加油站可以在循环的后半程走完。
class Solution:def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:#elsetotal_gas, curr_gas = 0, 0start_point = 0for i, element in enumerate(gas):curr_gas += element - cost[i]total_gas += element - cost[i]if curr_gas < 0:start_point = i + 1curr_gas = 0return start_point if total_gas >=0 else -1
近期做的最辛苦的一次，一开始思路被带偏了。
