描述
A bus has n stops numbered from 0 to n - 1 that form a circle. We know the distance between all pairs of neighboring stops where distance[i] is the distance between the stops number i and (i + 1) % n.
The bus goes along both directions i.e. clockwise and counterclockwise.
Return the shortest distance between the given start and destination stops.
Example 1:
Input: distance = [1,2,3,4], start = 0, destination = 1
Output: 1
Explanation: Distance between 0 and 1 is 1 or 9, minimum is 1.
Example 2:
Input: distance = [1,2,3,4], start = 0, destination = 2
Output: 3
Explanation: Distance between 0 and 2 is 3 or 7, minimum is 3.
Example 3:
Input: distance = [1,2,3,4], start = 0, destination = 3
Output: 4
Explanation: Distance between 0 and 3 is 6 or 4, minimum is 4.
Note:
1 <= n <= 10^4
distance.length == n
0 <= start, destination < n
0 <= distance[i] <= 10^4
解析
根据题意，就是在一个圈型的公交线上，找出两个站点的顺时针或者逆时针的最短距离，思路简单，就是先保证 start 在 destination 前面，然后比较顺时针 sum(distance[start:destination]) 的距离和和逆时针 sum(distance)-sum(distance[start:destination]) 的距离和（总距离-顺时针的距离和）的最小值为答案。
解答
class Solution(object):
def distanceBetweenBusStops(self, distance, start, destination):
"""
:type distance: List[int]
:type start: int
:type destination: int
:rtype: int
"""
if start>destination:
start, destination = destination, start
return min(sum(distance[start:destination]), sum(distance)-sum(distance[start:destination]))
运行结果
Runtime: 36 ms, faster than 26.83% of Python online submissions for Distance Between Bus Stops.
Memory Usage: 14.4 MB, less than 43.90% of Python online submissions for Distance Between Bus Stops.
原题链接：leetcode.com/problems/di…
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