? 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
1 package package01;
2
3 import edu.princeton.cs.algs4.StdIn;
4 import edu.princeton.cs.algs4.StdOut;
5
6 public class class01
7 {
8 private static final int BITS_PER_BYTE = 8;
9 private static final int BITS_PER_INT = 32;
10 private static final int R = 256;
11 private static final int CUTOFF = 15; // 尺寸不超过 CUTOFF 的数组使用插入排序
12
13 private class01() {}
14
15 public static void sort(String[] a) // 对字符串的排序
16 {
17 int n = a.length;
18 String[] aux = new String[n];
19 sortKernel(a, 0, n - 1, 0, aux);
20 }
21
22 private static void sortKernel(String[] a, int lo, int hi, int d, String[] aux)
23 {
24 if (hi <= lo + CUTOFF) // 数组较小，使用插入排序
25 {
26 insertion(a, lo, hi, d);
27 return;
28 }
29 int[] count = new int[R + 2];
30 for (int i = lo; i <= hi; i++)
31 count[charAt(a[i], d) + 2]++;
32 for (int r = 0; r < R + 1; r++)
33 count[r + 1] += count[r];
34 for (int i = lo; i <= hi; i++)
35 aux[count[charAt(a[i], d) + 1]++] = a[i];
36 for (int i = lo; i <= hi; i++) // 写回时注意偏移
37 a[i] = aux[i - lo];
38 for (int r = 0; r < R; r++) // 每个字母开头的子数组分别排序
39 sortKernel(a, lo + count[r], lo + count[r + 1] - 1, d + 1, aux);
40 }
41
42 private static void insertion(String[] a, int lo, int hi, int d) // 数组 a 关于第 d 位插入排序
43 {
44 for (int i = lo; i <= hi; i++)
45 {
46 for (int j = i; j > lo && less(a[j], a[j - 1], d); j--)
47 exch(a, j, j - 1);
48 }
49 }
50
51 private static void exch(String[] a, int i, int j)
52 {
53 String temp = a[i];
54 a[i] = a[j];
55 a[j] = temp;
56 }
57
58 private static int charAt(String s, int d) // 按索引取字符串的字符
59 {
60 assert d >= 0 && d <= s.length();
61 if (d == s.length())
62 return -1;
63 return s.charAt(d);
64 }
65
66 private static boolean less(String v, String w, int d)
67 {
68 //assert v.substring(0, d).equals(w.substring(0, d)); // 先检查等于
69 for (int i = d; i < Math.min(v.length(), w.length()); i++)
70 {
71 if (v.charAt(i) < w.charAt(i))
72 return true;
73 if (v.charAt(i) > w.charAt(i))
74 return false;
75 }
76 return v.length() < w.length();
77 }
78
79 public static void sort(int[] a) // 对数组的排序
80 {
81 int n = a.length;
82 int[] aux = new int[n];
83 sortKernel(a, 0, n - 1, 0, aux);
84 }
85
86 private static void sortKernel(int[] a, int lo, int hi, int d, int[] aux)
87 {
88 if (hi <= lo + CUTOFF)
89 {
90 insertion(a, lo, hi, d);
91 return;
92 }
93 int[] count = new int[R + 1];
94 int mask = R - 1;
95 int shift = BITS_PER_INT - BITS_PER_BYTE * d - BITS_PER_BYTE;
96 for (int i = lo; i <= hi; i++)
97 {
98 int c = (a[i] >> shift) & mask;
99 count[c + 1]++;
100 }
101 for (int r = 0; r < R; r++)
102 count[r + 1] += count[r];
103 if (d == 0) // 符号位，0x00-0x7F 要排在 0x80-0xFF 的后面
104 {
105 int shift1 = count[R] - count[R / 2];
106 int shift2 = count[R / 2];
107 for (int r = 0; r < R / 2; r++)
108 count[r] += shift1;
109 for (int r = R / 2; r < R; r++)
110 count[r] -= shift2;
111 }
112 for (int i = lo; i <= hi; i++)
113 {
114 int c = (a[i] >> shift) & mask;
115 aux[count[c]++] = a[i];
116 }
117 for (int i = lo; i <= hi; i++)
118 a[i] = aux[i - lo];
119 if (d == 4) // 已经到了最高位，不用分治了
120 return;
121 if (count[0] > 0)
122 sortKernel(a, lo, lo + count[0] - 1, d + 1, aux);
123 for (int r = 0; r < R; r++)
124 {
125 if (count[r + 1] > count[r]) // 存在数字第 d 位是 r 的数字需要排序
126 sortKernel(a, lo + count[r], lo + count[r + 1] - 1, d + 1, aux);
127 }
128 }
129
130 private static void insertion(int[] a, int lo, int hi, int d)
131 {
132 for (int i = lo; i <= hi; i++)
133 {
134 for (int j = i; j > lo && a[j] < a[j - 1]; j--)
135 exch(a, j, j - 1);
136 }
137 }
138
139 private static void exch(int[] a, int i, int j)
140 {
141 int temp = a[i];
142 a[i] = a[j];
143 a[j] = temp;
144 }
145
146 public static void main(String[] args)
147 {
148 String[] a = StdIn.readAllStrings();
149 int n = a.length;
150 sort(a);
151 for (int i = 0; i < n; i++)
152 StdOut.println(a[i]);
153 }
154 }
● 原地排序
1 package package01;
2
3 import edu.princeton.cs.algs4.StdIn;
4 import edu.princeton.cs.algs4.StdOut;
5
6 public class class01
7 {
8 private static final int R = 256;
9 private static final int CUTOFF = 0;
10
11 private class01() {}
12
13 public static void sort(String[] a)
14 {
15 int n = a.length;
16 sortKernel(a, 0, n - 1, 0);
17 }
18
19 private static void sortKernel(String[] a, int lo, int hi, int d)
20 {
21 if (hi <= lo + CUTOFF)
22 {
23 insertion(a, lo, hi, d);
24 return;
25 }
26 int[] heads = new int[R + 2]; // 需要记录每个字符出现的首位和末位
27 int[] tails = new int[R + 1];
28 for (int i = lo; i <= hi; i++)
29 heads[charAt(a[i], d) + 2]++;
30 heads[0] = lo; // heads 首位设为 lo，方便直接做 a 的索引
31 for (int r = 0; r < R + 1; r++)
32 {
33 heads[r + 1] += heads[r];
34 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
35 }
36 for (int r = 0; r < R + 1; r++) // 循环完成所有字符
37 {
38 for (; heads[r] < tails[r]; heads[r]++) // 循环直到所有 heads 向 tails 靠拢
39 {
40 // 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
41 for (int c = charAt(a[heads[r]], d); c + 1 != r; c = charAt(a[heads[r]], d))
42 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
43 }
44 }
45 for (int r = 0; r < R; r++) // 分治所有子数组
46 sortKernel(a, tails[r], tails[r + 1] - 1, d + 1);
47 }
48
49 private static void insertion(String[] a, int lo, int hi, int d)
50 {
51 for (int i = lo; i <= hi; i++)
52 {
53 for (int j = i; j > lo && less(a[j], a[j - 1], d); j--)
54 exch(a, j, j - 1);
55 }
56 }
57
58 private static void exch(String[] a, int i, int j)
59 {
60 String temp = a[i];
61 a[i] = a[j];
62 a[j] = temp;
63 }
64
65 private static int charAt(String s, int d)
66 {
67 assert d >= 0 && d <= s.length();
68 if (d == s.length())
69 return -1;
70 return s.charAt(d);
71 }
72
73 private static boolean less(String v, String w, int d)
74 {
75 // assert v.substring(0, d).equals(w.substring(0, d));
76 for (int i = d; i < Math.min(v.length(), w.length()); i++)
77 {
78 if (v.charAt(i) < w.charAt(i))
79 return true;
80 if (v.charAt(i) > w.charAt(i))
81 return false;
82 }
83 return v.length() < w.length();
84 }
85
86 public static void main(String[] args)
87 {
88 String[] a = StdIn.readAllStrings();
89 int n = a.length;
90 sort(a);
91 for (int i = 0; i < n; i++)
92 StdOut.println(a[i]);
93 }
94 }
● 三路快排
1 package package01;
2
3 import edu.princeton.cs.algs4.StdIn;
4 import edu.princeton.cs.algs4.StdOut;
5 import edu.princeton.cs.algs4.StdRandom;
6
7 public class class01
8 {
9 private static final int CUTOFF = 15;
10
11 private class01() {}
12
13 public static void sort(String[] a)
14 {
15 StdRandom.shuffle(a);
16 sortKernel(a, 0, a.length - 1, 0);
17 }
18
19 private static void sortKernel(String[] a, int lo, int hi, int d)
20 {
21 if (hi <= lo + CUTOFF)
22 {
23 insertion(a, lo, hi, d);
24 return;
25 }
26 int lt = lo, gt = hi;
27 int v = charAt(a[lo], d);
28 for (int i = lo + 1; i <= gt;)
29 {
30 int t = charAt(a[i], d);
31 if (t < v)
32 exch(a, lt++, i++);
33 else if (t > v)
34 exch(a, i, gt--);
35 else
36 i++;
37 }
38 sortKernel(a, lo, lt - 1, d);
39 if (v >= 0)
40 sortKernel(a, lt, gt, d + 1);
41 sortKernel(a, gt + 1, hi, d);
42 }
43
44 private static void insertion(String[] a, int lo, int hi, int d)
45 {
46 for (int i = lo; i <= hi; i++)
47 {
48 for (int j = i; j > lo && less(a[j], a[j - 1], d); j--)
49 exch(a, j, j - 1);
50 }
51 }
52
53 private static boolean less(String v, String w, int d)
54 {
55 assert v.substring(0, d).equals(w.substring(0, d));
56 for (int i = d; i < Math.min(v.length(), w.length()); i++)
57 {
58 if (v.charAt(i) < w.charAt(i))
59 return true;
60 if (v.charAt(i) > w.charAt(i))
61 return false;
62 }
63 return v.length() < w.length();
64 }
65
66 private static int charAt(String s, int d)
67 {
68 assert d >= 0 && d <= s.length();
69 if (d == s.length())
70 return -1;
71 return s.charAt(d);
72 }
73
74 private static void exch(String[] a, int i, int j)
75 {
76 String temp = a[i];
77 a[i] = a[j];
78 a[j] = temp;
79 }
80
81 public static void main(String[] args)
82 {
83 String[] a = StdIn.readAllStrings();
84 int n = a.length;
85
86 sort(a);
87
88 for (int i = 0; i < n; i++)
89 StdOut.println(a[i]);
90 }
91 }
原文地址：https://www.cnblogs.com/cuancuancuanhao/p/9869261.html
时间： 2024-08-04 21:32:29