[LeetCode] 1184. Distance Between Bus Stops

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:
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:
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:
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.

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

公交站间的距离。

环形公交路线上有 n 个站,按次序从 0 到 n - 1 进行编号。我们已知每一对相邻公交站之间的距离,distance[i] 表示编号为 i 的车站和编号为 (i + 1) % n 的车站之间的距离。

环线上的公交车都可以按顺时针和逆时针的方向行驶。

返回乘客从出发点 start 到目的地 destination 之间的最短距离。

来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/distance-between-bus-stops
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

思路

题目不难,但是做法很巧妙。题目要求计算的是两点之间的最短距离,最短距离的计算有两种方式,一种是顺时针,一种是逆时针。顺时针很好算,逆时针的算法是整圈的距离 - 顺时针的距离。

复杂度

时间O(n)
空间O(1)

代码

Java实现

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class Solution {
public int distanceBetweenBusStops(int[] distance, int start, int destination) {
int total = 0;
for (int dis : distance) {
total += dis;
}

if (start > destination) {
int temp = start;
start = destination;
destination = temp;
}
int dis1 = 0;
for (int i = start; i < destination; i++) {
dis1 += distance[i];
}
return Math.min(dis1, total - dis1);
}
}

[LeetCode] 1184. Distance Between Bus Stops
https://shurui91.github.io/posts/149749656.html
Author
Aaron Liu
Posted on
July 24, 2022
Licensed under