question:
Sublinear extra space. Develop a merge implementation that reduces the extra space requirement to max(M,N/M), based on the following idea: Divede the array into N/M blocks of size M (for simplicity in this description, assume that N is a multiple of M). Then, (i) considering the blocks as items with their first key as the sort key, sort them using selection sort; ans (ii) run through the array merging the first block with the second, then the second block with the third, and so forth.
answer:
//这道题不是很理解题目意思,我是先把这N/M个块先块内排完序,再把块两个两个归并成大块,再归并大块,直到排序完成
import edu.princeton.cs.algs4.*; public class Merge { public static void insertionsort(Comparable[] a, int lo, int hi)//插排数组a[]的lo到hi { for(int i = lo+1; i <= hi; i++) { for(int j = i; j > 0 && less(a[j], a[j-1]); j--) exch(a, j, j-1); } } private static void exch(Comparable[] a, int i, int j) { Comparable t = a[i]; a[i] = a[j]; a[j] = t; } private static Comparable[] aux; public static void merge(Comparable[] a, int lo, int mid, int hi) { int i = lo, j = mid + 1; for(int k = lo; k <= hi; k++)//复制 aux[k] = a[k]; for(int k = lo; k <= hi; k++)//把a[]填满就退出循环 { if(i > mid)//左侧已用完,则只能从右侧拿 a[k] = aux[j++]; else if(j > hi)//右侧已用完,则只能从左侧拿 a[k] = aux[i++]; else if(less(aux[i], aux[j]))//当前左侧比右侧小,从右侧拿 a[k] = aux[i++]; else//反之当前右侧比左侧小,从左侧拿 a[k] = aux[j++]; } } public static void sort(Comparable[] a) { int N = a.length; aux = new Comparable[N]; int M = 4;//我随便取了个M=4 for(int i = 0; i < N; i+=M)//分块插排 { insertionsort(a,i,Math.min(i+M-1,N-1));//这里取min就不用假设a的size是M的整数倍了 } for(int sz = M; sz < N; sz+=sz)//每次merge的大小size,从1开始到小于N的最大size for(int lo = 0; lo < N-sz; lo += sz+sz)//左sz和右sz进行归并,所以lo+=2sz merge(a, lo, lo+sz-1, Math.min(lo+sz+sz-1, N-1));//取min是怕最后边界的那个lo+sz+sz-1超过数组大小N-1,防止数组越界 } private static boolean less(Comparable v, Comparable w) { return v.compareTo(w) < 0; } private static void show(Comparable[] a) { for(int i = 0; i < a.length; i++) StdOut.print(a[i] + " "); StdOut.println(); } public static boolean isSorted(Comparable[] a) { for(int i = 1; i < a.length; i++) if(less(a[i], a[i-1])) return false; return true; } public static void main(String[] args) { String[] a = In.readStrings();//CTRL + d sort(a); assert isSorted(a); show(a); } }