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:



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

这题也很简单，我们遍历一次计算出总和，然后同时计算出给定范围的距离。这样我们就能得出哪个比较小了。

class DistanceBetweenBusStops : public Solution {
public:
    void Exec() {

    }

    int distanceBetweenBusStops(vector<int>& distance, int start, int destination) {
        int result = 0;
        if (start == destination) {
            return result;
        }
        if (start > destination) {
            std::swap(start, destination);
        }

        int sum = 0;
        for (int i = 0; i < distance.size(); i++) {
            sum += distance[i];
            if (i >= start && i < destination) {
                result += distance[i];
            }
        }

        return result < sum - result ? result : sum - result;
    }
};