问题：

给定n个汽车站之间的距离数组distances（相邻两个站相连，收尾站相连，可顺时针逆时针两个方向行驶）

求start站和destination站之间的最短距离。

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

        

 

 

解法：

从start站到destination站

两个方向行驶的距离 之和=总distance之和。

因此遍历所有distance

left=min(start, destination)

right=max(start, destination)

当 i 在 [left, right)（左闭右开区间）时，为顺时针行驶距离，

剩下距离求和，则为逆时针行驶距离。

最后返回二者较小值。

 

代码参考：

 1 class Solution {
 2 public:
 3     int distanceBetweenBusStops(vector<int>& distance, int start, int destination) {
 4         int res1=0;
 5         int res2=0;
 6         for(int i=0; i<distance.size(); i++){
 7             if(i>=min(start, destination) && i<max(start, destination)){
 8                 res1+=distance[i];
 9             }else{
10                 res2+=distance[i];
11             }
12         }
13         return min(res1, res2);
14     }
15 };