49. group anagramscopy for markdowncopy for markdown

Author: coderbean

.gitignore

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 *.tar.gz
 *.rar
 out/
+target/
 .idea/
 .DS_Store
 

LeetCode-Java.iml

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+<?xml version="1.0" encoding="UTF-8"?>
+<module org.jetbrains.idea.maven.project.MavenProjectsManager.isMavenModule="true" type="JAVA_MODULE" version="4">
+  <component name="NewModuleRootManager" LANGUAGE_LEVEL="JDK_1_8">
+    <output url="file://$MODULE_DIR$/target/classes" />
+    <output-test url="file://$MODULE_DIR$/target/test-classes" />
+    <content url="file://$MODULE_DIR$">
+      <sourceFolder url="file://$MODULE_DIR$/src/main/java" isTestSource="false" />
+      <excludeFolder url="file://$MODULE_DIR$/target" />
+    </content>
+    <orderEntry type="inheritedJdk" />
+    <orderEntry type="sourceFolder" forTests="false" />
+  </component>
+</module>

markdown/49. group anagramscopy for markdowncopy for markdown.md

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+### [49\. group anagramscopy for markdowncopy for markdown](https://leetcode.com/problems/group-anagrams/)
+
+difficulty: **medium**
+
+
+given an array of strings, group anagrams together.
+
+**example:**
+
+```
+input: ["eat", "tea", "tan", "ate", "nat", "bat"],
+output:
+[
+  ["ate","eat","tea"],
+  ["nat","tan"],
+  ["bat"]
+]```
+
+**note:**
+
+*   all inputs will be in lowercase.
+*   the order of your output does not matter.
+
+
+#### solution
+
+language: **java**
+
+```java
+class solution {
+    public list<list<string>> groupanagrams(string[] strs) {
+        list<list<string>> resultlist = new arraylist<>();
+        if (strs == null || strs.length == 0) {
+            return resultlist;
+        }
+        map<string,integer> groupsmap = new hashmap<>(); 
+        // 这里的 key 需要是 string，而不能使用数组，如果使用数组的话，key是内存地址，无法满足要求。
+        // value 存储的是这个组的字符串的 index
+        for (int i = 0; i < strs.length; i++) {
+            string str = strs[i];
+            char[] chars = str.tochararray();
+            arrays.sort(chars);
+            string s = string.valueof(chars); // 排序后得到最新的字符串
+            if (groupsmap.containskey(s)) {
+                resultlist.get(groupsmap.get(s)).add(str);
+                continue;
+            }
+            groupsmap.put(s,resultlist.size());
+            list<string> resultitem = new arraylist<>();
+            resultitem.add(str);
+            resultlist.add(resultitem);
+        }
+        return resultlist;
+    }
+}
+```
+![](http://ww4.sinaimg.cn/large/006tNc79ly1g4tvny2r5uj31b60qiq7l.jpg)

src/main/java/leetcode/_49_/Main.java

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+package leetcode._49_;
+
+import java.util.List;
+
+/**
+ * Created by zhangbo54 on 2019-03-04.
+ */
+public class Main {
+    public static void main(String[] args) {
+        Solution solution = new Solution();
+        String[] strs = {"eat", "tea", "tan", "ate", "nat", "bat"};
+        List<List<String>> lists = solution.groupAnagrams(strs);
+        System.out.println(lists);
+    }
+}
+

src/main/java/leetcode/_49_/Solution.java

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+package leetcode._49_;
+
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.HashMap;
+import java.util.List;
+import java.util.Map;
+
+class Solution {
+    public List<List<String>> groupAnagrams(String[] strs) {
+        List<List<String>> resultList = new ArrayList<>();
+        if (strs == null || strs.length == 0) {
+            return resultList;
+        }
+        Map<String,Integer> groupsMap = new HashMap<>();
+        // 这里的 key 需要是 String，而不能使用数组，如果使用数组的话，key是内存地址，无法满足要求。
+        // value 存储的是这个组的字符串的 index
+        for (int i = 0; i < strs.length; i++) {
+            String str = strs[i];
+            char[] chars = str.toCharArray();
+            Arrays.sort(chars);
+            String s = String.valueOf(chars); // 排序后得到最新的字符串
+            if (groupsMap.containsKey(s)) {
+                resultList.get(groupsMap.get(s)).add(str);
+                continue;
+            }
+            groupsMap.put(s,resultList.size());
+            List<String> resultItem = new ArrayList<>();
+            resultItem.add(str);
+            resultList.add(resultItem);
+        }
+        return resultList;
+    }
+}

src/main/java/leetcode/_49_/solution.md

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+### [49\. group anagramscopy for markdowncopy for markdown](https://leetcode.com/problems/group-anagrams/)
+
+difficulty: **medium**
+
+
+given an array of strings, group anagrams together.
+
+**example:**
+
+```
+input: ["eat", "tea", "tan", "ate", "nat", "bat"],
+output:
+[
+  ["ate","eat","tea"],
+  ["nat","tan"],
+  ["bat"]
+]```
+
+**note:**
+
+*   all inputs will be in lowercase.
+*   the order of your output does not matter.
+
+
+#### solution
+
+language: **java**
+
+```java
+class solution {
+    public list<list<string>> groupanagrams(string[] strs) {
+        list<list<string>> resultlist = new arraylist<>();
+        if (strs == null || strs.length == 0) {
+            return resultlist;
+        }
+        map<string,integer> groupsmap = new hashmap<>(); 
+        // 这里的 key 需要是 string，而不能使用数组，如果使用数组的话，key是内存地址，无法满足要求。
+        // value 存储的是这个组的字符串的 index
+        for (int i = 0; i < strs.length; i++) {
+            string str = strs[i];
+            char[] chars = str.tochararray();
+            arrays.sort(chars);
+            string s = string.valueof(chars); // 排序后得到最新的字符串
+            if (groupsmap.containskey(s)) {
+                resultlist.get(groupsmap.get(s)).add(str);
+                continue;
+            }
+            groupsmap.put(s,resultlist.size());
+            list<string> resultitem = new arraylist<>();
+            resultitem.add(str);
+            resultlist.add(resultitem);
+        }
+        return resultlist;
+    }
+}
+```
+![](http://ww4.sinaimg.cn/large/006tNc79ly1g4tvny2r5uj31b60qiq7l.jpg)