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算法（Algorithms）第4版 练习 2.2.11（2）
关键代码：
private static void sort(Comparable[] input, int lo, int hi) { if(lo >= hi)//just one entry in array return; int mid = lo + (hi-lo)/2; sort(input, lo, mid); sort(input, mid+1, hi); if(!less(input[mid+1],input[mid]))//input[mid+1] >= input[mid] return; merge(input, lo, mid, hi); }
整体：
package com.qiusongde; import edu.princeton.cs.algs4.In; import edu.princeton.cs.algs4.StdOut; public class MergeSkipMerge { private static Comparable[] aux; public static void sort(Comparable[] input) { int N = input.length; aux = new Comparable[N]; sort(input, 0, N-1); } private static void sort(Comparable[] input, int lo, int hi) { if(lo >= hi)//just one entry in array return; int mid = lo + (hi-lo)/2; sort(input, lo, mid); sort(input, mid+1, hi); if(!less(input[mid+1],input[mid]))//input[mid+1] >= input[mid] return; merge(input, lo, mid, hi); } private static void merge(Comparable[] input, int lo, int mid, int hi) { //copy input[lo,hi] to aux[lo,hi] for(int i = lo; i <= hi; i++) { aux[i] = input[i]; } int i = lo; int j = mid + 1; for(int k = lo; k <= hi; k++) { if(i > mid) input[k] = aux[j++]; else if(j > hi) input[k] = aux[i++]; else if(less(aux[j], aux[i])) input[k] = aux[j++]; else input[k] = aux[i++]; } StdOut.printf("merge(input, %4d, %4d, %4d)", lo, mid, hi); show(input);//for test } private static boolean less(Comparable v, Comparable w) { return v.compareTo(w) < 0; } private static void show(Comparable[] a) { //print the array, on a single line. 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) { //Read strings from standard input, sort them, and print. String[] input = In.readStrings(); show(input);//for test sort(input); assert isSorted(input); show(input);//for test } }
测试结果：
M E R G E S O R T E X A M P L E merge(input, 0, 0, 1)E M R G E S O R T E X A M P L E merge(input, 2, 2, 3)E M G R E S O R T E X A M P L E merge(input, 0, 1, 3)E G M R E S O R T E X A M P L E merge(input, 4, 5, 7)E G M R E O R S T E X A M P L E merge(input, 0, 3, 7)E E G M O R R S T E X A M P L E merge(input, 8, 8, 9)E E G M O R R S E T X A M P L E merge(input, 10, 10, 11)E E G M O R R S E T A X M P L E merge(input, 8, 9, 11)E E G M O R R S A E T X M P L E merge(input, 14, 14, 15)E E G M O R R S A E T X M P E L merge(input, 12, 13, 15)E E G M O R R S A E T X E L M P merge(input, 8, 11, 15)E E G M O R R S A E E L M P T X merge(input, 0, 7, 15)A E E E E G L M M O P R R S T X A E E E E G L M M O P R R S T X
性能对比：
For 20000 random Doubles 1000 trials
Merge is 3.6s MergeFasterM is 3.3s MergeUseInsert is 3.2s MergeSkipMerge is 3.4s
算法（Algorithms）第4版 练习 2.2.11（2）
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