技術標籤：陣列演算法技巧-字首和LeetCode

A bushas n stops numbered from 0 to n - 1 that forma circle. We know the distance between all pairs of neighboring stops where distance[i] is the distance between the stops numberi and (i + 1) % n.

The bus goes along both directionsi.e. clockwise and counterclockwise. Return the shortest distance between the givenstart

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

題意：環形公交路線上有 n 個站，按次序從 0 到 n - 1 進行編號，distance[i]

解法1 往兩邊掃描

class Solution {
public:
    int distanceBetweenBusStops(vector<int>& dist, int s, int d) {
        int n = dist.size(), left = dist[d], right = 0;
        for (int i = (s - 1 + n) % n; i != d; i = (i - 1 + n) % n) left += dist[i]; //往左走
        for (int i = s; i != d; i = (i + 1) % n) right += dist[i]; //往右走
        return min(left, right);
    }
};

執行效率如下：

執行用時：4 ms, 在所有 C++ 提交中擊敗了98.68% 的使用者
記憶體消耗：8.4 MB, 在所有 C++ 提交中擊敗了97.34% 的使用者

解法2 字首和

如果要多次查詢同一條環形公交車路線上的某兩個站點間最短距離，可以先使用字首和進行預處理：

class Solution {
public:
    int distanceBetweenBusStops(vector<int>& dist, int s, int d) {
        if (s > d) swap(s, d); //必不可少
        int n = dist.size();
        vector<int> psum(n + 1); //lsum[i]表示dist[0:i)區間和,即從0到i的距離
        for (int i = 0; i < n; ++i) psum[i + 1] = psum[i] + dist[i];
        int right = psum[d] - psum[s], left = psum[n] - right;
        return min(left, right);
    }
};

執行效率如下：

執行用時：8 ms, 在所有 C++ 提交中擊敗了68.42% 的使用者
記憶體消耗：8.6 MB, 在所有 C++ 提交中擊敗了93.36% 的使用者