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。

package week3;

import java.util.Arrays;
import edu.princeton.cs.algs4.StdRandom;

public class KthInTwoSortedArrays {
    /**
     * 寻找第K大元素
     * @param a 数组a
     * @param alo a的查找区间下界
     * @param ahi a的查找区间上界
     * @param b 数组b
     * @param blo b的查找区间下界
     * @param bhi b的查找区间上界
     * @param k 当前查找区间的第k大元素
     * @return
     */
    private static int find(int[] a, int alo, int ahi,int[] b, int blo, int bhi, int k){
        if(alo > ahi) return b[bhi-k+1];
        if(blo > bhi) return a[ahi-k+1];
        if(k==1) return a[ahi] > b[bhi]? a[ahi]:b[bhi];
        //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
        int bt = bhi-(k-1)/2 > blo ? bhi-(k-1)/2:blo;
        int at = ahi-(k-1)/2 > alo ? ahi-(k-1)/2:alo;
        if(a[at] >= b[bt]) 
            return find(a,alo,at-1,b,blo,bhi,k-(ahi-at+1));
        else
            return find(a,alo,ahi,b,blo,bt-1,k-(bhi-bt+1));
    }
    
    public static void main(String[] args){
        int n = 10;
        int n1 = StdRandom.uniform(n);
        int n2 = n-n1;
        int[] a = new int[n1];
        int[] b = new int[n2];
        for(int i=0;i<n1;i++){
            a[i] = StdRandom.uniform(100);
        }
        for(int i=0;i<n2;i++){
            b[i] = StdRandom.uniform(100);
        }
        Arrays.sort(a);
        Arrays.sort(b);
        System.out.println("a="+Arrays.toString(a));
        System.out.println("b="+Arrays.toString(b));
        int[] c = new int[n];
        int i = 0;
        int j = 0;
        int l = 0;
        while(l<n){
            if(i>=n1) c[l++] = b[j++];
            else if(j>=n2) c[l++] = a[i++];
            else{
                if(a[i] <= b[j])
                    c[l++] = a[i++];
                else
                    c[l++] = b[j++];
            }
                
        }
        System.out.println("c="+Arrays.toString(c));
        
        int k =StdRandom.uniform(1,n);
        int largestK = find(a,0,n1-1,b,0,n2-1,k);
        System.out.println("第"+k+"大元素是："+largestK);
    }
}