Coursera Algorithms week3 归并排序 练习测验1: Merging with smaller auxiliary array

题目原文：

Suppose that the subarray a[0] to a[n-1] is sorted and the subarray a[n] to a[2*n-1] is sorted. How can you merge the two subarrays so that a[0] to a[2*n-1] is sorted using an auxiliary array of length n (instead of 2n)

分析：

对两个大小分别为n的有序子数组进行归并，要求空间复杂度为n，正常情况下归并排序在此处的空间复杂度为2n，但是由于两个子数组分别是有序的，故用大小为n的额外子空间辅助归并是个很合理的要求，实现如下：

 1 import java.util.Arrays; 2 import edu.princeton.cs.algs4.StdRandom; 3  4 public class MergeSortedSubArray { 5   private static boolean less(Comparable v, Comparable w) { 6     return v.compareTo(w) < 0; 7   } 8   public static void merge(Comparable[] array){ 9     int n = array.length/2;10     Comparable[] aux = new Comparable[n];11     for(int i=0;i<n;i++){ //取左半边sorted的元素至辅助数组，因为未来归并左侧位置可能会被右侧元素占据12       aux[i] = array[i];13     }14     System.out.println(Arrays.toString(aux));15     int l = 0;16     int r = n;17     for(int k = 0; k<2*n;k++){18       if(l >= n) break;//辅助元素数组全部用完，array右侧不需要挪动位置了19       else if(r>=2*n) array[k]=aux[l++];//array原右侧元素全部放置合适位置，后面只需把辅助数组的元素挪到array右侧20       else if(less(array[r],aux[l])) array[k] = array[r++];21       else array[k] = aux[l++];22     }23   }24 25   public static void main(String[] args){26     int n = 10;27     int[] subarray1 = new int[n];28     int[] subarray2 = new int[n];29     for (int i = 0; i < n; i++) {30       subarray1[i] = StdRandom.uniform(100);31       subarray2[i] = StdRandom.uniform(100);32     }33     Arrays.sort(subarray1);34     Arrays.sort(subarray2);35     Integer[] array = new Integer[2*n];36     for(int i = 0; i<n;i++){37       array[i] = subarray1[i];38       array[n+i] = subarray2[i];39     }40     System.out.println(Arrays.toString(array));41     merge(array);42     System.out.println(Arrays.toString(array));43   }44 }

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原标题：Coursera Algorithms week3 归并排序 练习测验1: Merging with smaller auxiliary array

关键词：排序