A bushasnstops numbered from0ton - 1that forma circle. We know the distance between all pairs of neighboring stops wheredistance[i]is the distance between the stops numberiand(i + 1) % n.

The bus goes along both directionsi.e. clockwise and counterclockwise.

Return the shortest distance between the givenstart

anddestinationstops.

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

這道題說是有n個公交站形成了一個環，它們之間的距離用一個數組 distance 表示，其中 distance[i] 表示公交站i和 (i+1)%n

之間的距離。說是公交可以順時針和逆時針的開，問給定的任意起點和終點之間的最短距離。對於一道 Easy 題的身價，沒有太多的技巧而言，主要就是考察了一個迴圈陣列，求任意兩個點之間的距離，由於兩個方向都可以到達，那麼兩個方向的距離加起來就正好是整個陣列之和，所以只要求出一個方向的距離，另一個用總長度減去之前的距離就可以得到。所以這裡先求出所有數字之和，然後要求出其中一個方向的距離，由於處理迴圈陣列比較麻煩，所以這裡希望 start 小於 destination，可以從二者之間的較小值遍歷到較大值，累加距離之和，然後比較這個距離和，跟總距離減去該距離所得結果之間的較小值返回即可，參見程式碼如下：

解法一：

class Solution {
public:
    int distanceBetweenBusStops(vector<int>& distance, int start, int destination) {
        int total = accumulate(distance.begin(), distance.end(), 0), cur = 0, mx = max(start, destination);
        for (int i = min(start, destination); i < mx; ++i) {
            cur += distance[i];
        }
        return min(cur, total - cur);
    }
};

我們也可以只要一次遍歷就可以完成，因為最終都要是要遍歷所有的站點距離，當這個站點在 [start, destination) 範圍內，就累加到 sum1 中，否則就累加到 sum2 中。不過要要注意的是要確保 start 小於 destination，所以可以開始做個比較，若不滿足則交換二者。最終返回 sum1 和 sum2 中較小者即可，參見程式碼如下：

解法二：

class Solution {
public:
    int distanceBetweenBusStops(vector<int>& distance, int start, int destination) {
        int sum1 = 0, sum2 = 0, n = distance.size();
        if (start > destination) swap(start, destination);
        for (int i = 0; i < n; ++i) {
            if (i >= start && i < destination) {
                sum1 += distance[i];
            } else {
                sum2 += distance[i];
            }
        }
        return min(sum1, sum2);
    }
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1184

參考資料：

https://leetcode.com/problems/distance-between-bus-stops/

https://leetcode.com/problems/distance-between-bus-stops/discuss/377568/C%2B%2B-one-pass

https://leetcode.com/problems/distance-between-bus-stops/discuss/377444/Java-O(n)-solution-Easy-to-understand

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