文章目录
- 前缀树
- 桶排序
- 计数排序
- 基数排序
前缀树
实现一：
public class Code01_TrieTree {public static class Node1 {public int pass;public int end;public Node1[] nexts;// char tmp = 'b' (tmp - 'a')public Node1() {pass = 0;end = 0;// 0 a// 1 b// 2 c// .. ..// 25 z// nexts[i] == null i方向的路不存在// nexts[i] != null i方向的路存在nexts = new Node1[26];}}public static class Trie1 {private Node1 root;public Trie1() {root = new Node1();}public void insert(String word) {if (word == null) {return;}char[] str = word.toCharArray();Node1 node = root;node.pass++;int path = 0;for (int i = 0; i < str.length; i++) { // 从左往右遍历字符path = str[i] - 'a'; // 由字符，对应成走向哪条路if (node.nexts[path] == null) {node.nexts[path] = new Node1();}node = node.nexts[path];node.pass++;}node.end++;}public void delete(String word) {if (search(word) != 0) {char[] chs = word.toCharArray();Node1 node = root;node.pass--;int path = 0;for (int i = 0; i < chs.length; i++) {path = chs[i] - 'a';if (--node.nexts[path].pass == 0) {node.nexts[path] = null;return;}node = node.nexts[path];}node.end--;}}// word这个单词之前加入过几次public int search(String word) {if (word == null) {return 0;}char[] chs = word.toCharArray();Node1 node = root;int index = 0;for (int i = 0; i < chs.length; i++) {index = chs[i] - 'a';if (node.nexts[index] == null) {return 0;}node = node.nexts[index];}return node.end;}// 所有加入的字符串中，有几个是以pre这个字符串作为前缀的public int prefixNumber(String pre) {if (pre == null) {return 0;}char[] chs = pre.toCharArray();Node1 node = root;int index = 0;for (int i = 0; i < chs.length; i++) {index = chs[i] - 'a';if (node.nexts[index] == null) {return 0;}node = node.nexts[index];}return node.pass;}}
}
实现二：
public class Code02_TrieTree {public static class Node2 {public int pass;public int end;public HashMap<Integer, Node2> nexts;public Node2() {pass = 0;end = 0;nexts = new HashMap<>();}}public static class Trie2 {private Node2 root;public Trie2() {root = new Node2();}public void insert(String word) {if (word == null) {return;}char[] chs = word.toCharArray();Node2 node = root;node.pass++;int index = 0;for (int i = 0; i < chs.length; i++) {index = (int) chs[i];if (!node.nexts.containsKey(index)) {node.nexts.put(index, new Node2());}node = node.nexts.get(index);node.pass++;}node.end++;}public void delete(String word) {if (search(word) != 0) {char[] chs = word.toCharArray();Node2 node = root;node.pass--;int index = 0;for (int i = 0; i < chs.length; i++) {index = (int) chs[i];if (--node.nexts.get(index).pass == 0) {node.nexts.remove(index);return;}node = node.nexts.get(index);}node.end--;}}// word这个单词之前加入过几次public int search(String word) {if (word == null) {return 0;}char[] chs = word.toCharArray();Node2 node = root;int index = 0;for (int i = 0; i < chs.length; i++) {index = (int) chs[i];if (!node.nexts.containsKey(index)) {return 0;}node = node.nexts.get(index);}return node.end;}// 所有加入的字符串中，有几个是以pre这个字符串作为前缀的public int prefixNumber(String pre) {if (pre == null) {return 0;}char[] chs = pre.toCharArray();Node2 node = root;int index = 0;for (int i = 0; i < chs.length; i++) {index = (int) chs[i];if (!node.nexts.containsKey(index)) {return 0;}node = node.nexts.get(index);}return node.pass;}}
}
桶排序
计数排序
public class Code03_CountSort {// only for 0~200 valuepublic static void countSort(int[] arr) {if (arr == null || arr.length < 2) {return;}int max = Integer.MIN_VALUE;for (int i = 0; i < arr.length; i++) {max = Math.max(max, arr[i]);}int[] bucket = new int[max + 1];for (int i = 0; i < arr.length; i++) {bucket[arr[i]]++;}int i = 0;for (int j = 0; j < bucket.length; j++) {while (bucket[j]-- > 0) {arr[i++] = j;}}}
}
基数排序
public class Code04_RadixSort {public static void radixSort(int[] arr, int L, int R, int digit) {final int radix = 10;int i = 0, j = 0;// 有多少个数准备多少个辅助空间int[] help = new int[R - L + 1];for (int d = 1; d <= digit; d++) { // 有多少位就进出几次// 10个空间// count[0] 当前位(d位)是0的数字有多少个// count[1] 当前位(d位)是(0和1)的数字有多少个// count[2] 当前位(d位)是(0、1和2)的数字有多少个// count[i] 当前位(d位)是(0~i)的数字有多少个int[] count = new int[radix]; // count[0..9]for (i = L; i <= R; i++) {// 103 1 3// 209 1 9j = getDigit(arr[i], d);count[j]++;}for (i = 1; i < radix; i++) {count[i] = count[i] + count[i - 1];}for (i = R; i >= L; i--) {j = getDigit(arr[i], d);help[count[j] - 1] = arr[i];count[j]--;}for (i = L, j = 0; i <= R; i++, j++) {arr[i] = help[j];}}}public static int getDigit(int x, int d) {return ((x / ((int) Math.pow(10, d - 1))) % 10);}
}
