Leetcode平台上的Median of Two Sorted Arrays题目用Java快排实现

来源：互联网发布：自动弹钢琴的软件编辑：程序博客网时间：2024/06/05 13:31

Leetcode平台上的Median of Two Sorted Arrays题目，大意就是找两个已排序数组的中位数。今天先用快排的方式实现一下，代码如下：

There are two sorted arrays A and B of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

public class Solution {       public static int[] C;       public static void quickSort(int a[],int low, int high){        if(low < high){        int i = low;        int tmp;        for(int j = low; j< high; j++){            if(a[j]<=a[high]){                tmp = a[i];                a[i] = a[j];                a[j] = tmp;                ++ i;                }                       }        tmp = a[i];        a[i] = a[high];        a[high] = tmp;        quickSort(a,low,i-1);        quickSort(a,i+1,high);        }    }       public static double findMedianSortedArrays(int A[], int B[]) {        double i ;           //combine these two arrays into a new array C           int l = A.length+B.length;        C = new int[l];           System.arraycopy(A, 0, C, 0, A.length);           System.arraycopy(B, 0, C, A.length, B.length);                     quickSort(C,0,l-1);                   //for(int ii = 0; ii < l; ii ++)       //System.out.println(C[ii]);                     if(l%2 == 0)               i = (double) ((C[l/2]+C[l/2-1])/2.0);           else               i = (double) C[(l-1)/2];                     return i;     }    public static void main(String args[]){        int[] A = {1};        int[] B = {};        double median = findMedianSortedArrays(A,B);        System.out.println(median);    }}

快排递归深度logn，每一层的操作复杂度是O(n)所以复杂度应该是O(nlogn)。

对于快排能AC我表示相当意外，只能说leetcode平台给的测试用例不够强壮。如果给一个倒序排好序的数组，我这个快排就死掉了，相当于冒泡阿。

今天再用堆的思想实现一下，复杂度应该比这个快排低一些。

谢谢木易先森的指导！

0 0