Question

代码：

#include<bits/stdc++.h>

using namespace std;
void quicksort(int arr[], int low, int high) {
    if (high <= low)
        return;
    int i = low;
    int j = high + 1;
    int key = arr[low];
    while (true) {
        /*Find the number that bigger than key from left to right*/
        while (arr[++i] < key) {
            if (i == high) {
                break;
            }
        }
        /*Find the number that smaller than key from right to left*/
        while (arr[--j] > key) {
            if (j == low) {
                break;
            }
        }
        if (i >= j) {
            break;
        }
        /*exchange the number of i&j*/
        int temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
    }

    /*exchange the number*/
    int temp = arr[low];
    arr[low] = arr[j];
    arr[j] = temp;
    quicksort(arr, low, j - 1);
    quicksort(arr, j + 1, high);
}

int main() {
    int n;
    int a[1000010];
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    quicksort(a, 0, n - 1);
    for (int i = 0; i < n; i++) {
        cout << a[i] << " ";
    }
    return 0;
}

输入：

5
4 5 3 1 2

输出：

1 2 3 4 5

输出正常，我通过了3个测试点（共5个）。其他两个测试点说我用了太多时间。我的代码仍然不好吗？如何升级？

Answer 1

为枢轴选择中间值，从而消除了在内部循环中进行边界检查的需要。本示例使用两个函数，但是可以将它们组合为一个函数。逻辑是相同的：

const exampleMiddleware1 = async (ctx, next) => {
  return await next();

  console.log('Doing something');
};

const exampleMiddleware2 = async (ctx, next) => {
  next();

  console.log('Doing something');
};

const exampleMiddleware3 = async (ctx, next) => {
  return next();

  console.log('Doing something');
};

如何解决“ TLE”问题？我认为我的快速排序还可以

1 个答案: