Initial commit

Author: HUANGskysky

.idea/kotlinc.xml

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+<?xml version="1.0" encoding="UTF-8"?>
+<module type="JAVA_MODULE" version="4">
+  <component name="NewModuleRootManager" inherit-compiler-output="true">
+    <exclude-output />
+    <content url="file://$MODULE_DIR$">
+      <sourceFolder url="file://$MODULE_DIR$/src" isTestSource="false" />
+    </content>
+    <orderEntry type="inheritedJdk" />
+    <orderEntry type="sourceFolder" forTests="false" />
+  </component>
+</module>

.idea/misc.xml

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+package easy;
+
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.List;
+
+/**
+ * Created by Huangsky on 2018/5/25.
+ */
+public class NO438_FindAllAnagramsInAString {
+    public static void main(String[] args){
+        String s=new String("cbaebabacd");
+        String p=new String("abc");
+        List<Integer> list=findAnagrams(s,p);
+        System.out.println(list);
+        return ;
+    }
+    public static List<Integer> findAnagrams(String s, String p) {
+        List<Integer> list=new ArrayList<Integer>();
+        int len=p.length();
+        for(int i=0;i<=s.length()-len;i++){
+            String str=s.substring(i,i+len);
+            if(compareStr(str,p))
+                list.add(i);
+        }
+        return list;
+    }
+    public static boolean compareStr(String a,String b){
+        char[] strA=a.toCharArray();
+        char[] strB=b.toCharArray();
+        Arrays.sort(strA);
+        Arrays.sort(strB);
+        int i=0;
+        for(;i< strA.length;i++){
+            if (strA[i]!=strB[i])
+                break;
+        }
+        return i==strA.length;
+
+    }
+}

.idea/modules.xml

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+package easy;
+
+import java.util.LinkedList;
+import java.util.List;
+import java.util.Queue;
+
+/**
+ * Created by LXM on 2018/4/30.
+ */
+
+/**
+ * Description:
+ * Given a string S, we can transform every letter individually to be lowercase or uppercase to create another string.
+ * Return a list of all possible strings we could create.
+ *
+ * Example:
+ *
+     Input: S = "a1b2"
+     Output: ["a1b2", "a1B2", "A1b2", "A1B2"]
+
+     Input: S = "3z4"
+     Output: ["3z4", "3Z4"]
+
+     Input: S = "12345"
+     Output: ["12345"]
+ *
+ *
+ * Note:
+
+     S will be a string with length at most 12.
+     S will consist only of letters or digits.
+ */
+public class NO784_LetterCasePermutation {
+
+    public static void main(String[] args){
+        String s=new String("a1b2");
+        System.out.println("Result:::"+letterCasePermutation(s));
+    }
+
+    /**
+     * BFS-广度优先算法
+     */
+     public static List<String> letterCasePermutation(String S) {
+         Queue<String> queue=new LinkedList<>();
+         queue.add(S);
+         for(int i=0;i<S.length();i++){
+             int size=queue.size();
+             if(Character.isDigit(S.charAt(i)))
+                 continue;
+             for(int j=0;j<size;j++){
+                 String str=queue.poll();
+                 char[] arrCh=str.toCharArray();
+
+                 arrCh[i]=Character.toUpperCase(arrCh[i]);
+                 queue.add(new String(arrCh));
+
+                 arrCh[i]=Character.toLowerCase(arrCh[i]);
+                 queue.add(new String(arrCh));
+             }
+
+         }
+         return new LinkedList<>(queue);
+     }
+
+}

leetcode.iml

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+package easy;
+
+/**
+ * Created by Hsky on 2018/4/30.
+ */
+public class NO788_RotatedDigits {
+    /**
+     * Description:
+     *  X is a good number if after rotating each digit individually by 180 degrees, we get a valid number that is different from X.
+     *  Each digit must be rotated - we cannot choose to leave it alone.
+     *  A number is valid if each digit remains a digit after rotation. 0, 1, and 8 rotate to themselves; 2 and 5 rotate to each other;
+     *  6 and 9 rotate to each other, and the rest of the numbers do not rotate to any other number and become invalid.
+     *  Now given a positive number N, how many numbers X from 1 to N are good?
+     *
+     *  Example:
+     *
+         Input: 10
+         Output: 4
+         Explanation:
+         There are four good numbers in the range [1, 10] : 2, 5, 6, 9.
+         Note that 1 and 10 are not good numbers, since they remain unchanged after rotating.
+
+
+     *  Note:
+         N  will be in range [1, 10000].
+     */
+
+    public static void  main(String[] args){
+
+        System.out.println("result:"+rotatedDigits(100));
+    }
+    public static int rotatedDigits(int N) {
+        int count = 0;
+        for (int i = 1; i <= N; i ++) {
+            if (isValid(i)) count ++;
+        }
+        return count;
+    }
+
+    public static boolean isValid(int N) {
+        /*
+         Valid if N contains ATLEAST ONE 2, 5, 6, 9
+         AND NO 3, 4 or 7s
+         */
+        boolean validFound = false;
+        while (N > 0) {
+            if (N % 10 == 2) validFound = true;
+            if (N % 10 == 5) validFound = true;
+            if (N % 10 == 6) validFound = true;
+            if (N % 10 == 9) validFound = true;
+            if (N % 10 == 3) return false;
+            if (N % 10 == 4) return false;
+            if (N % 10 == 7) return false;
+            N = N / 10;
+        }
+        return validFound;
+    }
+}

src/easy/NO438_FindAllAnagramsInAString.java

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+package easy;
+
+import java.util.*;
+
+/**
+ * Created by Hsky on 2018/4/30.
+ */
+public class NO819_MostCommonWord {
+    /**
+     * Description:
+     * Given a paragraph and a list of banned words, return the most frequent word that is not in the list of banned words.
+     * It is guaranteed there is at least one word that isn't banned, and that the answer is unique.
+     * Words in the list of banned words are given in lowercase, and free of punctuation.  Words in the paragraph are not case sensitive.
+     * The answer is in lowercase.
+     * <p>
+     * Example:
+     * Example:
+     * Input:
+     * paragraph = "Bob hit a ball, the hit BALL flew far after it was hit."
+     * banned = ["hit"]
+     * Output: "ball"
+     * Explanation:
+     * "hit" occurs 3 times, but it is a banned word.
+     * "ball" occurs twice (and no other word does), so it is the most frequent non-banned word in the paragraph.
+     * Note that words in the paragraph are not case sensitive,
+     * that punctuation is ignored (even if adjacent to words, such as "ball,"),
+     * and that "hit" isn't the answer even though it occurs more because it is banned.
+     * <p>
+     * <p>
+     * Note:
+     * <p>
+     * 1 <= paragraph.length <= 1000.
+     * 1 <= banned.length <= 100.
+     * 1 <= banned[i].length <= 10.
+     * The answer is unique, and written in lowercase (even if its occurrences in paragraph may have uppercase symbols, and even if it is a proper noun.)
+     * paragraph only consists of letters, spaces, or the punctuation symbols !?',;.
+     * Different words in paragraph are always separated by a space.
+     * There are no hyphens or hyphenated words.
+     * Words only consist of letters, never apostrophes or other punctuation symbols.
+     */
+//
+    public static void main(String[] args) {
+        String paragraph=new String("Bob!");//hit a ball, the hit BALL flew far after it was hit.
+        String[] banned=new String[]{"hit"};
+        String res=mostCommonWord(paragraph,banned);
+        System.out.println(res);
+
+    }
+
+    public static String mostCommonWord(String p, String[] banned) {
+        Set<String> ban = new HashSet<>(Arrays.asList(banned));
+        Map<String, Integer> count = new HashMap<>();
+        String[] words = p.replaceAll("\\pP" , "").toLowerCase().split("\\s+");
+        String res = "";
+        int max = 0;
+        for (String w : words) {
+            if (!ban.contains(w)) {
+                count.put(w, count.getOrDefault(w, 0) + 1);
+                if (count.get(w) > max) {
+                    res = w;
+                    max = count.get(w);
+                }}}
+        return res;
+    }
+}