1184 Distance Between Bus Stops 公交站间的距离
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:
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
题目描述:
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。
示例 :
示例 1：
输入：distance = [1,2,3,4], start = 0, destination = 1
输出：1
解释：公交站 0 和 1 之间的距离是 1 或 9，最小值是 1。
示例 2：
输入：distance = [1,2,3,4], start = 0, destination = 2
输出：3
解释：公交站 0 和 2 之间的距离是 3 或 7，最小值是 3。
示例 3：
输入：distance = [1,2,3,4], start = 0, destination = 3
输出：4
解释：公交站 0 和 3 之间的距离是 6 或 4，最小值是 4。
提示：
1 <= n <= 10^4
distance.length == n
0 <= start, destination < n
0 <= distance[i] <= 10^4
思路:
由于公交车只能顺时针或者逆时针移动, 只要比较两个方向的距离和即可
时间复杂度O(n), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
int distanceBetweenBusStops(vector<int>& distance, int start, int destination)
{
int clockwise = 0, sum = accumulate(distance.begin(), distance.end(), 0);
if (start > destination) swap(start, destination);
for (int i = start; i < destination; ++i) clockwise += distance[i];
return min(clockwise, sum - clockwise);
}
};
Java:
class Solution {
public int distanceBetweenBusStops(int[] distance, int start, int destination) {
int clockwise = 0, sum = 0;
if (start > destination) {
start ^= destination;
destination ^= start;
start ^= destination;
}
for (int i : distance) sum += i;
for (int i = start; i < destination; ++i) clockwise += distance[i];
return Math.min(clockwise, sum - clockwise);
}
}
Python:
class Solution:
def distanceBetweenBusStops(self, distance: List[int], start: int, destination: int) -> int:
return min(sum(distance[start:destination]), sum(distance) - sum(distance[start:destination])) if start < destination else min(sum(distance[destination:start]), sum(distance) - sum(distance[destination:start]))
