《算法》第五章部分程序 part 2

▶ 书中第五章部分程序，包括在加上自己补充的代码，字符串高位优先排序（计数 + 插排），（原地排序），（三路快排，与前面的三路归并排序相同）

● 计数 + 插排

 package package01;

 import edu.princeton.cs.algs4.StdIn;
 import edu.princeton.cs.algs4.StdOut;

 public class class01
 {
     private static final int BITS_PER_BYTE = 8;
     private static final int BITS_PER_INT = 32;
     private static final int R = 256;
     private static final int CUTOFF = 15;   // 尺寸不超过 CUTOFF 的数组使用插入排序

     private class01() {}

     public static void sort(String[] a)     // 对字符串的排序
     {
         int n = a.length;
         String[] aux = new String[n];
         sortKernel(a, 0, n - 1, 0, aux);
     }

     private static void sortKernel(String[] a, int lo, int hi, int d, String[] aux)
     {
         if (hi <= lo + CUTOFF)              // 数组较小，使用插入排序
         {
             insertion(a, lo, hi, d);
             return;
         }
         int[] count = new int[R + 2];
         for (int i = lo; i <= hi; i++)
             count[charAt(a[i], d) + 2]++;
         for (int r = 0; r < R + 1; r++)
             count[r + 1] += count[r];
         for (int i = lo; i <= hi; i++)
             aux[count[charAt(a[i], d) + 1]++] = a[i];
         for (int i = lo; i <= hi; i++)      // 写回时注意偏移
             a[i] = aux[i - lo];
         for (int r = 0; r < R; r++)         // 每个字母开头的子数组分别排序
             sortKernel(a, lo + count[r], lo + count[r + 1] - 1, d + 1, aux);
     }

     private static void insertion(String[] a, int lo, int hi, int d)    // 数组 a 关于第 d 位插入排序
     {
         for (int i = lo; i <= hi; i++)
         {
             for (int j = i; j > lo && less(a[j], a[j - 1], d); j--)
                 exch(a, j, j - 1);
         }
     }

     private static void exch(String[] a, int i, int j)
     {
         String temp = a[i];
         a[i] = a[j];
         a[j] = temp;
     }

     private static int charAt(String s, int d)              // 按索引取字符串的字符
     {
         assert d >= 0 && d <= s.length();
         if (d == s.length())
             return -1;
         return s.charAt(d);
     }

     private static boolean less(String v, String w, int d)
     {
         //assert v.substring(0, d).equals(w.substring(0, d)); // 先检查等于
         for (int i = d; i < Math.min(v.length(), w.length()); i++)
         {
             if (v.charAt(i) < w.charAt(i))
                 return true;
             if (v.charAt(i) > w.charAt(i))
                 return false;
         }
         return v.length() < w.length();
     }

     public static void sort(int[] a)        // 对数组的排序
     {
         int n = a.length;
         int[] aux = new int[n];
         sortKernel(a, 0, n - 1, 0, aux);
     }

     private static void sortKernel(int[] a, int lo, int hi, int d, int[] aux)
     {
         if (hi <= lo + CUTOFF)
         {
             insertion(a, lo, hi, d);
             return;
         }
         int[] count = new int[R + 1];
         int mask = R - 1;
         int shift = BITS_PER_INT - BITS_PER_BYTE * d - BITS_PER_BYTE;
         for (int i = lo; i <= hi; i++)
         {
             int c = (a[i] >> shift) & mask;
             count[c + 1]++;
         }
         for (int r = 0; r < R; r++)
             count[r + 1] += count[r];
         if (d == 0)                         // 符号位，0x00-0x7F 要排在 0x80-0xFF 的后面
         {
             int shift1 = count[R] - count[R / 2];
             int shift2 = count[R / 2];
             for (int r = 0; r < R / 2; r++)
                 count[r] += shift1;
             for (int r = R / 2; r < R; r++)
                 count[r] -= shift2;
         }
         for (int i = lo; i <= hi; i++)
         {
             int c = (a[i] >> shift) & mask;
             aux[count[c]++] = a[i];
         }
         for (int i = lo; i <= hi; i++)
             a[i] = aux[i - lo];
         if (d == 4)                         // 已经到了最高位，不用分治了
             return;
         if (count[0] > 0)
             sortKernel(a, lo, lo + count[0] - 1, d + 1, aux);
         for (int r = 0; r < R; r++)
         {
             if (count[r + 1] > count[r])    // 存在数字第 d 位是 r 的数字需要排序
                 sortKernel(a, lo + count[r], lo + count[r + 1] - 1, d + 1, aux);
         }
     }

     private static void insertion(int[] a, int lo, int hi, int d)
     {
         for (int i = lo; i <= hi; i++)
         {
             for (int j = i; j > lo && a[j] < a[j - 1]; j--)
                 exch(a, j, j - 1);
         }
     }

     private static void exch(int[] a, int i, int j)
     {
         int temp = a[i];
         a[i] = a[j];
         a[j] = temp;
     }

     public static void main(String[] args)
     {
         String[] a = StdIn.readAllStrings();
         int n = a.length;
         sort(a);
         for (int i = 0; i < n; i++)
             StdOut.println(a[i]);
     }
 }

● 原地排序

 package package01;

 import edu.princeton.cs.algs4.StdIn;
 import edu.princeton.cs.algs4.StdOut;

 public class class01
 {
     private static final int R = 256;
     private static final int CUTOFF = 0;

     private class01() {}

     public static void sort(String[] a)
     {
         int n = a.length;
         sortKernel(a, 0, n - 1, 0);
     }

     private static void sortKernel(String[] a, int lo, int hi, int d)
     {
         if (hi <= lo + CUTOFF)
         {
             insertion(a, lo, hi, d);
             return;
         }
         int[] heads = new int[R + 2];       // 需要记录每个字符出现的首位和末位
         int[] tails = new int[R + 1];
         for (int i = lo; i <= hi; i++)
             heads[charAt(a[i], d) + 2]++;
         heads[0] = lo;                      // heads 首位设为 lo，方便直接做 a 的索引
         for (int r = 0; r < R + 1; r++)
         {
             heads[r + 1] += heads[r];
             tails[r] = heads[r + 1];        // tails 等于前缀和的 heads 左移一格，等于下一个字符出现的位置
         }
         for (int r = 0; r < R + 1; r++)     // 循环完成所有字符
         {
             for (; heads[r] < tails[r]; heads[r]++)     // 循环直到所有 heads 向 tails 靠拢
             {
                 // a[heads[r]] 作为临时位置，每次把其上字符串发送到排序后正确的位置上，替换下来的字符串放回，用于下一次发送
                 for (int c = charAt(a[heads[r]], d); c + 1 != r; c = charAt(a[heads[r]], d))
                     exch(a, heads[r], heads[c + 1]++);  // 每次发送后要后移同一字符串的目标位置
             }
         }
         for (int r = 0; r < R; r++)                     // 分治所有子数组
             sortKernel(a, tails[r], tails[r + 1] - 1, d + 1);
     }

     private static void insertion(String[] a, int lo, int hi, int d)
     {
         for (int i = lo; i <= hi; i++)
         {
             for (int j = i; j > lo && less(a[j], a[j - 1], d); j--)
                 exch(a, j, j - 1);
         }
     }

     private static void exch(String[] a, int i, int j)
     {
         String temp = a[i];
         a[i] = a[j];
         a[j] = temp;
     }

     private static int charAt(String s, int d)
     {
         assert d >= 0 && d <= s.length();
         if (d == s.length())
             return -1;
         return s.charAt(d);
     }

     private static boolean less(String v, String w, int d)
     {
         // assert v.substring(0, d).equals(w.substring(0, d));
         for (int i = d; i < Math.min(v.length(), w.length()); i++)
         {
             if (v.charAt(i) < w.charAt(i))
                 return true;
             if (v.charAt(i) > w.charAt(i))
                 return false;
         }
         return v.length() < w.length();
     }

     public static void main(String[] args)
     {
         String[] a = StdIn.readAllStrings();
         int n = a.length;
         sort(a);
         for (int i = 0; i < n; i++)
             StdOut.println(a[i]);
     }
 }

● 三路快排

 package package01;

 import edu.princeton.cs.algs4.StdIn;
 import edu.princeton.cs.algs4.StdOut;
 import edu.princeton.cs.algs4.StdRandom;

 public class class01
 {
     private static final int CUTOFF = 15;

     private class01() {}

     public static void sort(String[] a)
     {
         StdRandom.shuffle(a);
         sortKernel(a, 0, a.length - 1, 0);
     }

     private static void sortKernel(String[] a, int lo, int hi, int d)
     {
         if (hi <= lo + CUTOFF)
         {
             insertion(a, lo, hi, d);
             return;
         }
         int lt = lo, gt = hi;
         int v = charAt(a[lo], d);
         for (int i = lo + 1; i <= gt;)
         {
             int t = charAt(a[i], d);
             if (t < v)
                 exch(a, lt++, i++);
             else if (t > v)
                 exch(a, i, gt--);
             else
                 i++;
         }
         sortKernel(a, lo, lt - 1, d);
         if (v >= 0)
             sortKernel(a, lt, gt, d + 1);
         sortKernel(a, gt + 1, hi, d);
     }

     private static void insertion(String[] a, int lo, int hi, int d)
     {
         for (int i = lo; i <= hi; i++)
         {
             for (int j = i; j > lo && less(a[j], a[j - 1], d); j--)
                 exch(a, j, j - 1);
         }
     }

     private static boolean less(String v, String w, int d)
     {
         assert v.substring(0, d).equals(w.substring(0, d));
         for (int i = d; i < Math.min(v.length(), w.length()); i++)
         {
             if (v.charAt(i) < w.charAt(i))
                 return true;
             if (v.charAt(i) > w.charAt(i))
                 return false;
         }
         return v.length() < w.length();
     }

     private static int charAt(String s, int d)
     {
         assert d >= 0 && d <= s.length();
         if (d == s.length())
             return -1;
         return s.charAt(d);
     }

     private static void exch(String[] a, int i, int j)
     {
         String temp = a[i];
         a[i] = a[j];
         a[j] = temp;
     }

     public static void main(String[] args)
     {
         String[] a = StdIn.readAllStrings();
         int n = a.length;

         sort(a);

         for (int i = 0; i < n; i++)
             StdOut.println(a[i]);
     }
 }