[leetcode] 1184. Distance Between Bus Stops

[leetcode] 1184. Distance Between Bus Stops

Description

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 = 1Output: 1Explanation: Distance between 0 and 1 is 1 or 9, minimum is 1.

Example 2:

Input: distance = [1,2,3,4], start = 0, destination = 2Output: 3Explanation: Distance between 0 and 2 is 3 or 7, minimum is 3.

Example 3:

Input: distance = [1,2,3,4], start = 0, destination = 3Output: 4Explanation: Distance between 0 and 3 is 6 or 4, minimum is 4.

Constraints:

1 <= n <= 10^4.distance.length == n.0 <= start, destination < n0 <= distance[i] <= 10^4

分析

题目的意思是：给定一个数组，求出顺时针遍历和逆时针遍历求和的最小值。我这里用了两次循环来模拟顺时针遍历和逆时针遍历，看了别人的参考发现start和destination可以先比较大小，保持start值小于destination的值然后进行遍历。anyway，我也实现了，时间复杂度O(n)，还不错

代码

class Solution: def distanceBetweenBusStops(self, distance: List[int], start: int, destination: int) -> int: res=0 n=len(distance) t=0 for i in range(start,n+destination): j=i%n if(j==destination): break t+=distance[j] res=t t=0 for i in range(destination,n+start): j=i%n if(j==start): break t+=distance[j] if(t>res): return res return t

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