modify code

MCA算法突击课/第03期/mca_08/Code01_Trie.java

@@ -0,0 +1,102 @@
+package 第03期.mca_08;
+
+public class Code01_Trie {
+
+	// 测试链接 : https://leetcode.cn/problems/implement-trie-ii-prefix-tree/
+	// 提交Trie类可以直接通过
+	// 原来代码是对的，但是既然找到了直接测试的链接，那就直接测吧
+	// 这个链接上要求实现的功能和课上讲的完全一样
+	// 该前缀树的路用数组实现
+	class Trie {
+
+		class Node {
+			public int pass;
+			public int end;
+			public Node[] nexts;
+
+			public Node() {
+				pass = 0;
+				end = 0;
+				nexts = new Node[26];
+			}
+		}
+
+		private Node root;
+
+		public Trie() {
+			root = new Node();
+		}
+
+		public void insert(String word) {
+			if (word == null) {
+				return;
+			}
+			char[] str = word.toCharArray();
+			Node 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 Node();
+				}
+				node = node.nexts[path];
+				node.pass++;
+			}
+			node.end++;
+		}
+
+		public void erase(String word) {
+			if (countWordsEqualTo(word) != 0) {
+				char[] chs = word.toCharArray();
+				Node 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--;
+			}
+		}
+
+		public int countWordsEqualTo(String word) {
+			if (word == null) {
+				return 0;
+			}
+			char[] chs = word.toCharArray();
+			Node 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;
+		}
+
+		public int countWordsStartingWith(String pre) {
+			if (pre == null) {
+				return 0;
+			}
+			char[] chs = pre.toCharArray();
+			Node 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;
+		}
+	}
+
+}

MCA算法突击课/第03期/mca_08/Code02_Trie.java

@@ -0,0 +1,104 @@
+package 第03期.mca_08;
+
+import java.util.HashMap;
+
+public class Code02_Trie {
+
+	// 测试链接 : https://leetcode.cn/problems/implement-trie-ii-prefix-tree/
+	// 提交Trie类可以直接通过
+	// 原来代码是对的，但是既然找到了直接测试的链接，那就直接测吧
+	// 这个链接上要求实现的功能和课上讲的完全一样
+	// 该前缀树的路用哈希表实现
+	class Trie {
+
+		class Node {
+			public int pass;
+			public int end;
+			public HashMap<Integer, Node> nexts;
+
+			public Node() {
+				pass = 0;
+				end = 0;
+				nexts = new HashMap<>();
+			}
+		}
+
+		private Node root;
+
+		public Trie() {
+			root = new Node();
+		}
+
+		public void insert(String word) {
+			if (word == null) {
+				return;
+			}
+			char[] chs = word.toCharArray();
+			Node 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 Node());
+				}
+				node = node.nexts.get(index);
+				node.pass++;
+			}
+			node.end++;
+		}
+
+		public void erase(String word) {
+			if (countWordsEqualTo(word) != 0) {
+				char[] chs = word.toCharArray();
+				Node 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--;
+			}
+		}
+
+		public int countWordsEqualTo(String word) {
+			if (word == null) {
+				return 0;
+			}
+			char[] chs = word.toCharArray();
+			Node 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;
+		}
+
+		public int countWordsStartingWith(String pre) {
+			if (pre == null) {
+				return 0;
+			}
+			char[] chs = pre.toCharArray();
+			Node 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;
+		}
+	}
+
+}

MCA算法突击课/第03期/mca_08/Code03_PrefixAndSuffixSearch.java

@@ -0,0 +1,111 @@
+package 第03期.mca_08;
+
+import java.util.ArrayList;
+
+// 设计一个包含一些单词的特殊词典，并能够通过前缀和后缀来检索单词。
+// 实现 WordFilter 类：
+// WordFilter(string[] words) 使用词典中的单词 words 初始化对象
+// f(string pref, string suff) 
+// 返回词典中具有前缀 prefix 和后缀 suff 的单词的下标
+// 如果存在不止一个满足要求的下标，返回其中 最大的下标
+// 如果不存在这样的单词，返回 -1 。
+// 测试链接 : https://leetcode.cn/problems/prefix-and-suffix-search/
+public class Code03_PrefixAndSuffixSearch {
+
+	// 提交以下这个类
+	class WordFilter {
+
+		class TrieNode {
+			TrieNode[] nexts;
+			ArrayList<Integer> indies;
+
+			public TrieNode() {
+				nexts = new TrieNode[26];
+				indies = new ArrayList<>();
+			}
+		}
+
+		TrieNode preHead;
+
+		TrieNode sufHead;
+
+		public WordFilter(String[] words) {
+			preHead = new TrieNode();
+			sufHead = new TrieNode();
+			for (int i = 0; i < words.length; i++) {
+				String word = words[i];
+				TrieNode cur = preHead;
+				for (int j = 0; j < word.length(); j++) {
+					int path = word.charAt(j) - 'a';
+					if (cur.nexts[path] == null) {
+						cur.nexts[path] = new TrieNode();
+					}
+					cur = cur.nexts[path];
+					cur.indies.add(i);
+				}
+				cur = sufHead;
+				for (int j = word.length() - 1; j >= 0; j--) {
+					int path = word.charAt(j) - 'a';
+					if (cur.nexts[path] == null) {
+						cur.nexts[path] = new TrieNode();
+					}
+					cur = cur.nexts[path];
+					cur.indies.add(i);
+				}
+			}
+		}
+
+		public int f(String pref, String suff) {
+			ArrayList<Integer> preList = null;
+			TrieNode cur = preHead;
+			for (int i = 0; i < pref.length() && cur != null; i++) {
+				cur = cur.nexts[pref.charAt(i) - 'a'];
+			}
+			if (cur != null) {
+				preList = cur.indies;
+			}
+			if (preList == null) {
+				return -1;
+			}
+			ArrayList<Integer> sufList = null;
+			cur = sufHead;
+			for (int i = suff.length() - 1; i >= 0 && cur != null; i--) {
+				cur = cur.nexts[suff.charAt(i) - 'a'];
+			}
+			if (cur != null) {
+				sufList = cur.indies;
+			}
+			if (sufList == null) {
+				return -1;
+			}
+			ArrayList<Integer> small = preList.size() <= sufList.size() ? preList : sufList;
+			ArrayList<Integer> big = small == preList ? sufList : preList;
+			for (int i = small.size() - 1; i >= 0; i--) {
+				if (bs(big, small.get(i))) {
+					return small.get(i);
+				}
+			}
+			return -1;
+		}
+
+		private boolean bs(ArrayList<Integer> sorted, int num) {
+			int l = 0;
+			int r = sorted.size() - 1;
+			int m = 0;
+			int midValue = 0;
+			while (l <= r) {
+				m = (l + r) / 2;
+				midValue = sorted.get(m);
+				if (midValue == num) {
+					return true;
+				} else if (midValue < num) {
+					l = m + 1;
+				} else {
+					r = m - 1;
+				}
+			}
+			return false;
+		}
+	}
+
+}

MCA算法突击课/第03期/mca_08/Code04_SameTeams.java

@@ -0,0 +1,62 @@
+package 第03期.mca_08;
+
+import java.util.ArrayList;
+import java.util.HashMap;
+
+// 来自字节飞书团队
+// 假设数组a和数组b为两组信号
+// 1) length(b) <= length(a)
+// 2) 对于任意0<=i<length(b), 有b[i+1] - b[i] == a[i+1] - a[i]
+// 那么就称信号b和信号a一致，记为b==a
+// 给你好多b数组，假设有m个: b0数组、b1数组...
+// 给你好多a数组，假设有n个: a0数组、a1数组...
+// 返回一个长度为m的结果数组ans，ans[i]表示 : bi数组和多少个a数组一致
+public class Code04_SameTeams {
+
+	public static class TrieNode {
+		public ArrayList<Integer> indices;
+		public HashMap<Integer, TrieNode> nexts;
+
+		public TrieNode() {
+			indices = new ArrayList<>();
+			nexts = new HashMap<>();
+		}
+
+	}
+
+	public static int[] sameTeamsArray(int[][] bs, int[][] as) {
+		int m = bs.length;
+		TrieNode root = new TrieNode();
+		TrieNode cur = null;
+		for (int i = 0; i < m; i++) {
+			int k = bs[i].length;
+			cur = root;
+			for (int j = 1; j < k; j++) {
+				int diff = bs[i][j] - bs[i][j - 1];
+				if (!cur.nexts.containsKey(diff)) {
+					cur.nexts.put(diff, new TrieNode());
+				}
+				cur = cur.nexts.get(diff);
+			}
+			cur.indices.add(i);
+		}
+		int[] ans = new int[m];
+		int n = as.length;
+		for (int i = 0; i < n; i++) {
+			int k = as[i].length;
+			cur = root;
+			for (int j = 1; j < k; j++) {
+				int diff = as[i][j] - as[i][j - 1];
+				if (!cur.nexts.containsKey(diff)) {
+					break;
+				}
+				cur = cur.nexts.get(diff);
+				for (int index : cur.indices) {
+					ans[index]++;
+				}
+			}
+		}
+		return ans;
+	}
+
+}

MCA算法突击课/第03期/mca_08/Code05_MaximumFrequencyStack.java

@@ -1,4 +1,4 @@
-package 第03期.mca_07;
+package 第03期.mca_08;
 
 import java.util.ArrayList;
 import java.util.HashMap;

MCA算法突击课/第03期/mca_08/Code06_InsertDeleteGetRandom.java

@@ -1,4 +1,4 @@
-package 第03期.mca_07;
+package 第03期.mca_08;
 
 import java.util.HashMap;
 

MCA算法突击课/第03期/mca_08/Code07_ExpressionCompute.java

@@ -0,0 +1,57 @@
+package 第03期.mca_08;
+
+import java.util.LinkedList;
+
+// 本题测试链接 : https://leetcode.com/problems/basic-calculator-iii/
+public class Code07_ExpressionCompute {
+
+	public static int calculate(String str) {
+		return f(str.toCharArray(), 0)[0];
+	}
+
+	// 请从str[i...]往下算，遇到字符串终止位置或者右括号，就停止
+	// 返回两个值，长度为2的数组
+	// 0) 负责的这一段的结果是多少
+	// 1) 负责的这一段计算到了哪个位置
+	public static int[] f(char[] str, int i) {
+		LinkedList<String> queue = new LinkedList<String>();
+		int cur = 0;
+		int[] bra = null;
+		// 从i出发，开始撸串
+		while (i < str.length && str[i] != ')') {
+			if (str[i] >= '0' && str[i] <= '9') {
+				cur = cur * 10 + str[i++] - '0';
+			} else if (str[i] != '(') { // 遇到的是运算符号
+				addNum(queue, cur, str[i++]);
+				cur = 0;
+			} else { // 遇到左括号了
+				bra = f(str, i + 1);
+				cur = bra[0];
+				i = bra[1] + 1;
+			}
+		}
+		addNum(queue, cur, '+');
+		return new int[] { getAns(queue), i };
+	}
+
+	public static void addNum(LinkedList<String> queue, int num, char op) {
+		if (!queue.isEmpty() && (queue.peekLast().equals("*") || queue.peekLast().equals("/"))) {
+			String top = queue.pollLast();
+			int pre = Integer.valueOf(queue.pollLast());
+			num = top.equals("*") ? (pre * num) : (pre / num);
+		}
+		queue.addLast(String.valueOf(num));
+		queue.addLast(String.valueOf(op));
+	}
+
+	public static int getAns(LinkedList<String> queue) {
+		int ans = Integer.valueOf(queue.pollFirst());
+		while (queue.size() > 1) {
+			String op = queue.pollFirst();
+			int num = Integer.valueOf(queue.pollFirst());
+			ans += op.equals("+") ? num : -num;
+		}
+		return ans;
+	}
+
+}