init

Author: douhuatong

.gitignore

@@ -1,27 +1,9 @@
-# ---> macOS
-.DS_Store
-.AppleDouble
-.LSOverride
-
-# Icon must end with two \r
-Icon
-
-
-# Thumbnails
-._*
-
-# Files that might appear in the root of a volume
-.DocumentRevisions-V100
-.fseventsd
-.Spotlight-V100
-.TemporaryItems
-.Trashes
-.VolumeIcon.icns
-
-# Directories potentially created on remote AFP share
-.AppleDB
-.AppleDesktop
-Network Trash Folder
-Temporary Items
-.apdisk
+# Created by .ignore support plugin (hsz.mobi)
+### Example user template template
+### Example user template
 
+# IntelliJ project files
+.idea
+*.iml
+out
+gen

pom.xml

@@ -0,0 +1,40 @@
+<?xml version="1.0" encoding="UTF-8"?>
+<project xmlns="http://maven.apache.org/POM/4.0.0"
+         xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
+         xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd">
+    <modelVersion>4.0.0</modelVersion>
+
+    <groupId>com.leetcode</groupId>
+    <artifactId>leetcode</artifactId>
+    <version>1.0-SNAPSHOT</version>
+
+
+    <properties>
+        <project.build.sourceEncoding>UTF-8</project.build.sourceEncoding>
+        <project.reporting.outputEncoding>UTF-8</project.reporting.outputEncoding>
+        <java.version>1.8</java.version>
+    </properties>
+    <dependencies>
+        <dependency>
+            <groupId>org.apache.commons</groupId>
+            <artifactId>commons-lang3</artifactId>
+            <version>3.9</version>
+        </dependency>
+    </dependencies>
+
+
+    <build>
+        <plugins>
+            <plugin>
+                <groupId>org.apache.maven.plugins</groupId>
+                <artifactId>maven-compiler-plugin</artifactId>
+                <configuration>
+                    <source>${java.version}</source>
+                    <target>${java.version}</target>
+                    <encoding>${project.build.sourceEncoding}</encoding>
+                    <showWarnings>true</showWarnings>
+                </configuration>
+            </plugin>
+        </plugins>
+    </build>
+</project>

src/main/java/com/leetcode/DecodeString.java

@@ -0,0 +1,87 @@
+package com.leetcode;
+
+import java.util.Collections;
+import java.util.LinkedList;
+import java.util.Stack;
+
+/**
+ * 给定一个经过编码的字符串，返回它解码后的字符串。
+ * <p>
+ * 编码规则为: k[encoded_string]，表示其中方括号内部的 encoded_string 正好重复 k 次。注意 k 保证为正整数。
+ * <p>
+ * 你可以认为输入字符串总是有效的；输入字符串中没有额外的空格，且输入的方括号总是符合格式要求的。
+ * <p>
+ * 此外，你可以认为原始数据不包含数字，所有的数字只表示重复的次数 k ，例如不会出现像 3a 或 2[4] 的输入。
+ * <p>
+ * 示例:
+ * <p>
+ * s = "3[a]2[bc]", 返回 "aaabcbc".
+ * s = "3[a2[c]]", 返回 "accaccacc".
+ * s = "2[abc]3[cd]ef", 返回 "abcabccdcdcdef".
+ * <p>
+ * 来源：力扣（LeetCode）
+ * 链接：https://leetcode-cn.com/problems/decode-string
+ * 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
+ */
+public class DecodeString {
+
+    int ptr;
+
+    public String decodeString(String s) {
+        Stack<String> stk = new Stack<>();
+        ptr = 0;
+        while (ptr < s.length()) {
+            char cur = s.charAt(ptr);
+            if (Character.isDigit(cur)) {//是否为数字
+                // 获取一个数字并进栈
+                String digits = getDigits(s);
+                stk.add(digits);
+            } else if (Character.isLetter(cur) || cur == '[') {
+                stk.add(String.valueOf(s.charAt(ptr++)));
+            } else {
+                ++ptr;
+                Stack<String> sub = new Stack<>();
+                while (!"[".equals(stk.peek())) {
+                    sub.add(stk.pop());
+                }
+                Collections.reverse(sub);
+                // 左括号出栈
+                stk.pop();
+                // 此时栈顶为当前 sub 对应的字符串应该出现的次数
+                int repTime = Integer.parseInt(stk.pop());
+                StringBuffer t = new StringBuffer();
+                String o = getString(sub);
+                // 构造字符串
+                while (repTime-- > 0) {
+                    t.append(o);
+                }
+                // 将构造好的字符串入栈
+                stk.add(t.toString());
+            }
+        }
+        return getString(stk);
+    }
+
+    public String getString(Stack<String> v) {
+        StringBuffer ret = new StringBuffer();
+        for (String s : v) {
+            ret.append(s);
+        }
+        return ret.toString();
+    }
+
+
+    public String getDigits(String s) {
+        StringBuffer ret = new StringBuffer();
+        while (Character.isDigit(s.charAt(ptr))) {
+            ret.append(s.charAt(ptr++));
+        }
+        return ret.toString();
+    }
+
+
+    public static void main(String[] args) {
+        DecodeString decodeString = new DecodeString();
+        System.err.println(decodeString.decodeString("3[a]2[bc]"));
+    }
+}

src/main/java/com/leetcode/SubarraysDivByK.java

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+package com.leetcode;
+
+
+import java.util.Arrays;
+import java.util.HashMap;
+import java.util.Map;
+
+/**
+ * 给定一个整数数组 A，返回其中元素之和可被 K 整除的（连续、非空）子数组的数目。
+ * <p>
+ * 示例：
+ * <p>
+ * 输入：A = [4,5,0,-2,-3,1], K = 5
+ * 输出：7
+ * 解释：
+ * 有 7 个子数组满足其元素之和可被 K = 5 整除：
+ * [4, 5, 0, -2, -3, 1], [5], [5, 0], [5, 0, -2, -3], [0], [0, -2, -3], [-2, -3]
+ * <p>
+ * 来源：力扣（LeetCode）
+ * 链接：https://leetcode-cn.com/problems/subarray-sums-divisible-by-k
+ * 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
+ */
+public class SubarraysDivByK {
+
+    public static int subarraysDivByK(int[] A, int K) {
+        int length = A.length;
+        int r = 0;
+        for (int i = 0; i <= length; i++) {
+            for (int j = 0; j <= length; j++) {
+                if (j < i) {
+                    int newArray[] = Arrays.copyOfRange(A, j, i);
+                    if (newArray.length > 0) {
+                        int sum = 0;
+                        for (int item : newArray) {
+                            sum = sum + item;
+                        }
+                        if (sum % K == 0) {
+                            r++;
+                        }
+                    }
+                }
+            }
+        }
+        return r;
+    }
+
+
+    public static int subarraysDivByK2(int[] A, int K) {
+        Map<Integer, Integer> record = new HashMap<>();
+        record.put(0, 1);
+        int sum = 0, ans = 0;
+        for (int elem : A) {
+            sum += elem;
+            int modulus = (sum % K + K) % K;
+            int same = record.getOrDefault(modulus, 0);
+            ans = ans + same;
+            record.put(modulus, same + 1);
+        }
+        return ans;
+    }
+
+    public static void main(String[] args) {
+        int[] a = {4, 5, 0, -2, -3, 1};
+        System.err.println(subarraysDivByK2(a, 5));
+    }
+}