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  • leetcode134 Gas Station

     1 """
     2 There are N gas stations along a circular route, where the amount of gas at station i is gas[i].
     3 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.
     4 Return the starting gas station's index if you can travel around the circuit once in the clockwise direction, otherwise return -1.
     5 Note:
     6     If there exists a solution, it is guaranteed to be unique.
     7     Both input arrays are non-empty and have the same length.
     8     Each element in the input arrays is a non-negative integer.
     9 Example 1:
    10 Input:
    11 gas  = [1,2,3,4,5]
    12 cost = [3,4,5,1,2]
    13 Output: 3
    14 Explanation:
    15 Start at station 3 (index 3) and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
    16 Travel to station 4. Your tank = 4 - 1 + 5 = 8
    17 Travel to station 0. Your tank = 8 - 2 + 1 = 7
    18 Travel to station 1. Your tank = 7 - 3 + 2 = 6
    19 Travel to station 2. Your tank = 6 - 4 + 3 = 5
    20 Travel to station 3. The cost is 5. Your gas is just enough to travel back to station 3.
    21 Therefore, return 3 as the starting index.
    22 Example 2:
    23 Input:
    24 gas  = [2,3,4]
    25 cost = [3,4,3]
    26 Output: -1
    27 Explanation:
    28 You can't start at station 0 or 1, as there is not enough gas to travel to the next station.
    29 Let's start at station 2 and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
    30 Travel to station 0. Your tank = 4 - 3 + 2 = 3
    31 Travel to station 1. Your tank = 3 - 3 + 3 = 3
    32 You cannot travel back to station 2, as it requires 4 unit of gas but you only have 3.
    33 Therefore, you can't travel around the circuit once no matter where you start.
    34 """
    35 """
    36 解法一:自己AC
    37 按照题目要求正常实现
    38 """
    39 class Solution1:
    40     def canCompleteCircuit(self, gas, cost):
    41         cur = 0
    42         for i in range(len(gas)):
    43             cur = gas[i]
    44             for j in range(i, i+len(gas)):
    45                 cur -= cost[j % len(gas)]
    46                 if cur < 0:
    47                     break
    48                 cur += gas[(j+1) % len(gas)]
    49             if cur >= 0:
    50                 return i
    51         return -1
    52 """
    53 解法二:算法思想,主要遵从两个原则:
    54 1.出发点必须满足gas>cost,否则不可能出发
    55 2.总的gas>cost,否则一定走不完
    56 """
    57 class Solution:
    58     def canCompleteCircuit(self, gas, cost):
    59         total = 0
    60         res = 0
    61         cur = 0
    62         for i in range(len(gas)):
    63             cur += gas[i] - cost[i]
    64             if cur < 0:
    65                 res = i + 1
    66                 total += cur
    67                 cur = 0
    68         return res if (total + cur) >= 0 else -1
    69 if __name__ == '__main__':
    70     gas = [1, 2, 3, 4, 5]
    71     cost = [3, 4, 5, 1, 2]
    72     ans = Solution1()
    73     print(ans.canCompleteCircuit(gas, cost))
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  • 原文地址:https://www.cnblogs.com/yawenw/p/12466640.html
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