标签：boolean 有序 pos break 额外 排序 port random bar
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 }
Coursera Algorithms week3 归并排序 练习测验1: Merging with smaller auxiliary array
标签：boolean 有序 pos break 额外 排序 port random bar
原文地址：http://www.cnblogs.com/evasean/p/7220112.html
