Coursera Algorithms week3 快速排序 练习测验: Selection in two sorted arrays（从两个有序数组中寻找第K大元素）...

题目原文

Selection in two sorted arrays. Given two sorted arrays a[] and b[], of sizes n1 and n2, respectively, design an algorithm to find the k^th largest key. The order  of growth of the worst case running time of your algorithm should be logn, where n = n1 + n2.

Version 1 : n1 = n2 and k = n/2

Version 2: k = n/2

Version 3: no restrictions

分析：

 这个题目是要求从两个有序数组中寻找第K大元素，直观解法就是把连个数组归并排序，这样时间复杂度就是O(n)，但是题目只要求知道第K大元素，其余比K大的元素的排列顺序并不关心。

默认数组a和数组b都是从小到大排序的。不论第K大元素在a或b中，其右侧元素一定大于K，可考虑分别从a和b中排除右侧(k-1)/2个元素，剩下的就是在未排除区间寻找第(k-已排除元素)大元素。通过递归逐步缩小比较范围。具体算法见如下代码。

 1 package week3;
 2 
 3 import java.util.Arrays;
 4 import edu.princeton.cs.algs4.StdRandom;
 5 
 6 public class KthInTwoSortedArrays {
 7     /**
 8      * 寻找第K大元素
 9      * @param a 数组a
10      * @param alo a的查找区间下界
11      * @param ahi a的查找区间上界
12      * @param b 数组b
13      * @param blo b的查找区间下界
14      * @param bhi b的查找区间上界
15      * @param k 当前查找区间的第k大元素
16      * @return
17      */
18     private static int find(int[] a, int alo, int ahi,int[] b, int blo, int bhi, int k){
19         if(alo > ahi) return b[bhi-k+1];
20         if(blo > bhi) return a[ahi-k+1];
21         if(k==1) return a[ahi] > b[bhi]? a[ahi]:b[bhi];
22         //直接用bhi-(k-1)/2或ahi-(k-1)/2进行查找可能会将第K个元素给漏掉
23         int bt = bhi-(k-1)/2 > blo ? bhi-(k-1)/2:blo;
24         int at = ahi-(k-1)/2 > alo ? ahi-(k-1)/2:alo;
25         if(a[at] >= b[bt]) 
26             return find(a,alo,at-1,b,blo,bhi,k-(ahi-at+1));
27         else
28             return find(a,alo,ahi,b,blo,bt-1,k-(bhi-bt+1));
29     }
30     
31     public static void main(String[] args){
32         int n = 10;
33         int n1 = StdRandom.uniform(n);
34         int n2 = n-n1;
35         int[] a = new int[n1];
36         int[] b = new int[n2];
37         for(int i=0;i){
38             a[i] = StdRandom.uniform(100);
39         }
40         for(int i=0;i){
41             b[i] = StdRandom.uniform(100);
42         }
43         Arrays.sort(a);
44         Arrays.sort(b);
45         System.out.println("a="+Arrays.toString(a));
46         System.out.println("b="+Arrays.toString(b));
47         int[] c = new int[n];
48         int i = 0;
49         int j = 0;
50         int l = 0;
51         while(l<n){
52             if(i>=n1) c[l++] = b[j++];
53             else if(j>=n2) c[l++] = a[i++];
54             else{
55                 if(a[i] <= b[j])
56                     c[l++] = a[i++];
57                 else
58                     c[l++] = b[j++];
59             }
60                 
61         }
62         System.out.println("c="+Arrays.toString(c));
63         
64         int k =StdRandom.uniform(1,n);
65         int largestK = find(a,0,n1-1,b,0,n2-1,k);
66         System.out.println("第"+k+"大元素是："+largestK);
67     }
68 }

 

转载于:https://www.cnblogs.com/evasean/p/7263160.html