pull/6/head
左程云 4 years ago
parent 745f7e0535
commit 0147f549e1

@ -0,0 +1,77 @@
package class32;
public class Code01_IndexTree {
public static class IndexTree {
private int[] tree;
private int N;
public IndexTree(int size) {
N = size;
tree = new int[N + 1];
}
public int sum(int index) {
int ret = 0;
while (index > 0) {
ret += tree[index];
index -= index & -index;
}
return ret;
}
public void add(int index, int d) {
while (index <= N) {
tree[index] += d;
index += index & -index;
}
}
}
public static class Right {
private int[] nums;
private int N;
public Right(int size) {
N = size + 1;
nums = new int[N + 1];
}
public int sum(int index) {
int ret = 0;
for (int i = 1; i <= index; i++) {
ret += nums[i];
}
return ret;
}
public void add(int index, int d) {
nums[index] += d;
}
}
public static void main(String[] args) {
int N = 100;
int V = 100;
int testTime = 2000000;
IndexTree tree = new IndexTree(N);
Right test = new Right(N);
System.out.println("test begin");
for (int i = 0; i < testTime; i++) {
int index = (int) (Math.random() * N) + 1;
if (Math.random() <= 0.5) {
int add = (int) (Math.random() * V);
tree.add(index, add);
test.add(index, add);
} else {
if (tree.sum(index) != test.sum(index)) {
System.out.println("Oops!");
}
}
}
System.out.println("test finish");
}
}

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package class32;
// 测试链接https://leetcode.com/problems/range-sum-query-2d-mutable
// 但这个题是付费题目
// 提交时把类名、构造函数名从Code02_IndexTree2D改成NumMatrix
public class Code02_IndexTree2D {
private int[][] tree;
private int[][] nums;
private int N;
private int M;
public Code02_IndexTree2D(int[][] matrix) {
if (matrix.length == 0 || matrix[0].length == 0) {
return;
}
N = matrix.length;
M = matrix[0].length;
tree = new int[N + 1][M + 1];
nums = new int[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++) {
update(i, j, matrix[i][j]);
}
}
}
private int sum(int row, int col) {
int sum = 0;
for (int i = row + 1; i > 0; i -= i & (-i)) {
for (int j = col + 1; j > 0; j -= j & (-j)) {
sum += tree[i][j];
}
}
return sum;
}
public void update(int row, int col, int val) {
if (N == 0 || M == 0) {
return;
}
int add = val - nums[row][col];
nums[row][col] = val;
for (int i = row + 1; i <= N; i += i & (-i)) {
for (int j = col + 1; j <= M; j += j & (-j)) {
tree[i][j] += add;
}
}
}
public int sumRegion(int row1, int col1, int row2, int col2) {
if (N == 0 || M == 0) {
return 0;
}
return sum(row2, col2) + sum(row1 - 1, col1 - 1) - sum(row1 - 1, col2) - sum(row2, col1 - 1);
}
}

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package class32;
import java.util.LinkedList;
import java.util.Queue;
public class Code03_AC1 {
public static class Node {
public int end; // 有多少个字符串以该节点结尾
public Node fail;
public Node[] nexts;
public Node() {
end = 0;
fail = null;
nexts = new Node[26];
}
}
public static class ACAutomation {
private Node root;
public ACAutomation() {
root = new Node();
}
// 你有多少个匹配串就调用多少次insert
public void insert(String s) {
char[] str = s.toCharArray();
Node cur = root;
int index = 0;
for (int i = 0; i < str.length; i++) {
index = str[i] - 'a';
if (cur.nexts[index] == null) {
Node next = new Node();
cur.nexts[index] = next;
}
cur = cur.nexts[index];
}
cur.end++;
}
public void build() {
Queue<Node> queue = new LinkedList<>();
queue.add(root);
Node cur = null;
Node cfail = null;
while (!queue.isEmpty()) {
cur = queue.poll(); // 父
for (int i = 0; i < 26; i++) { // 下级所有的路
if (cur.nexts[i] != null) { // 该路下有子节点
cur.nexts[i].fail = root; // 初始时先设置一个值
cfail = cur.fail;
while (cfail != null) { // cur不是头节点
if (cfail.nexts[i] != null) {
cur.nexts[i].fail = cfail.nexts[i];
break;
}
cfail = cfail.fail;
}
queue.add(cur.nexts[i]);
}
}
}
}
public int containNum(String content) {
char[] str = content.toCharArray();
Node cur = root;
Node follow = null;
int index = 0;
int ans = 0;
for (int i = 0; i < str.length; i++) {
index = str[i] - 'a';
while (cur.nexts[index] == null && cur != root) {
cur = cur.fail;
}
cur = cur.nexts[index] != null ? cur.nexts[index] : root;
follow = cur;
while (follow != root) {
if (follow.end == -1) {
break;
}
{ // 不同的需求,在这一段{ }之间修改
ans += follow.end;
follow.end = -1;
} // 不同的需求,在这一段{ }之间修改
follow = follow.fail;
}
}
return ans;
}
}
public static void main(String[] args) {
ACAutomation ac = new ACAutomation();
ac.insert("dhe");
ac.insert("he");
ac.insert("c");
ac.build();
System.out.println(ac.containNum("cdhe"));
}
}

@ -0,0 +1,126 @@
package class32;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
public class Code04_AC2 {
// 前缀树的节点
public static class Node {
// 如果一个nodeend为空不是结尾
// 如果end不为空表示这个点是某个字符串的结尾end的值就是这个字符串
public String end;
// 只有在上面的end变量不为空的时候endUse才有意义
// 表示,这个字符串之前有没有加入过答案
public boolean endUse;
public Node fail;
public Node[] nexts;
public Node() {
endUse = false;
end = null;
fail = null;
nexts = new Node[26];
}
}
public static class ACAutomation {
private Node root;
public ACAutomation() {
root = new Node();
}
public void insert(String s) {
char[] str = s.toCharArray();
Node cur = root;
int index = 0;
for (int i = 0; i < str.length; i++) {
index = str[i] - 'a';
if (cur.nexts[index] == null) {
Node next = new Node();
cur.nexts[index] = next;
}
cur = cur.nexts[index];
}
cur.end = s;
}
public void build() {
Queue<Node> queue = new LinkedList<>();
queue.add(root);
Node cur = null;
Node cfail = null;
while (!queue.isEmpty()) {
// 当前节点弹出,
// 当前节点的所有后代加入到队列里去,
// 当前节点给它的子去设置fail指针
// cur -> 父亲
cur = queue.poll();
for (int i = 0; i < 26; i++) { // 所有的路
if (cur.nexts[i] != null) { // 找到所有有效的路
cur.nexts[i].fail = root; //
cfail = cur.fail;
while (cfail != null) {
if (cfail.nexts[i] != null) {
cur.nexts[i].fail = cfail.nexts[i];
break;
}
cfail = cfail.fail;
}
queue.add(cur.nexts[i]);
}
}
}
}
public List<String> containWords(String content) {
char[] str = content.toCharArray();
Node cur = root;
Node follow = null;
int index = 0;
List<String> ans = new ArrayList<>();
for (int i = 0; i < str.length; i++) {
index = str[i] - 'a'; // 路
// 如果当前字符在这条路上没配出来就随着fail方向走向下条路径
while (cur.nexts[index] == null && cur != root) {
cur = cur.fail;
}
// 1) 现在来到的路径,是可以继续匹配的
// 2) 现在来到的节点,就是前缀树的根节点
cur = cur.nexts[index] != null ? cur.nexts[index] : root;
follow = cur;
while (follow != root) {
if(follow.endUse) {
break;
}
// 不同的需求,在这一段之间修改
if (follow.end != null) {
ans.add(follow.end);
follow.endUse = true;
}
// 不同的需求,在这一段之间修改
follow = follow.fail;
}
}
return ans;
}
}
public static void main(String[] args) {
ACAutomation ac = new ACAutomation();
ac.insert("dhe");
ac.insert("he");
ac.insert("abcdheks");
// 设置fail指针
ac.build();
List<String> contains = ac.containWords("abcdhekskdjfafhasldkflskdjhwqaeruv");
for (String word : contains) {
System.out.println(word);
}
}
}
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