mirror of
https://github.com/halfrost/LeetCode-Go.git
synced 2026-03-13 10:02:05 +08:00
82 lines
2.6 KiB
Markdown
Executable File
82 lines
2.6 KiB
Markdown
Executable File
# [1184. Distance Between Bus Stops](https://leetcode.com/problems/distance-between-bus-stops/)
|
||
|
||
## 题目
|
||
|
||
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.
|
||
|
||
**Constraints:**
|
||
|
||
- `1 <= n <= 10^4`
|
||
- `distance.length == n`
|
||
- `0 <= start, destination < n`
|
||
- `0 <= distance[i] <= 10^4`
|
||
|
||
|
||
## 题目大意
|
||
|
||
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。
|
||
|
||
提示：
|
||
|
||
- 1 <= n <= 10^4
|
||
- distance.length == n
|
||
- 0 <= start, destination < n
|
||
- 0 <= distance[i] <= 10^4
|
||
|
||
|
||
## 解题思路
|
||
|
||
|
||
- 给出一个数组，代表的是公交车站每站直接的距离。距离是按照数组下标的顺序给出的，公交车可以按照顺时针行驶，也可以按照逆时针行驶。问行驶的最短距离是多少。
|
||
- 按照题意，分别算出顺时针和逆时针的行驶距离，比较两者距离，取出小值就是结果。
|
||
|
||
|
||
## 代码
|
||
|
||
```go
|
||
|
||
package leetcode
|
||
|
||
func distanceBetweenBusStops(distance []int, start int, destination int) int {
|
||
clockwiseDis, counterclockwiseDis, n := 0, 0, len(distance)
|
||
for i := start; i != destination; i = (i + 1) % n {
|
||
clockwiseDis += distance[i]
|
||
}
|
||
for i := destination; i != start; i = (i + 1) % n {
|
||
counterclockwiseDis += distance[i]
|
||
}
|
||
if clockwiseDis < counterclockwiseDis {
|
||
return clockwiseDis
|
||
}
|
||
return counterclockwiseDis
|
||
}
|
||
|
||
``` |