feat: add 049

Author: Blankj

README.md

@@ -65,6 +65,7 @@
 |19|[Remove Nth Node From End of List][019]|Linked List, Two Pointers|
 |33|[Search in Rotated Sorted Array][033]|Arrays, Binary Search|
 |43|[Multiply Strings][043]|Math, String|
+|49|[Group Anagrams][049]|Hash Table, String|
 |554|[Brick Wall][554]|Hash Table|
 
 
@@ -127,6 +128,7 @@
 [019]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/019/README.md
 [033]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/033/README.md
 [043]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/043/README.md
+[049]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/049/README.md
 [554]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/554/README.md
 
 [004]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/004/README.md

note/049/README.md

@@ -0,0 +1,60 @@
+# [Group Anagrams][title]
+
+## Description
+
+Given an array of strings, group anagrams together.
+
+For example, given: `["eat", "tea", "tan", "ate", "nat", "bat"]`, 
+Return:
+
+```
+[
+  ["ate", "eat","tea"],
+  ["nat","tan"],
+  ["bat"]
+]
+```
+
+**Note:** All inputs will be in lower-case.
+
+**Tags:** Hash Table, String
+
+
+## 思路0
+
+题意是给你一组字符串，让你把其中同位异构字符串分组，同位异构字符串就是组成字符串的字符都相同，但是字符放的位置可以不同。既然要分组，那么关键就是如何确定他们是同位异构字符串呢，想到的自然就是对其排序，排序之后他们就都是同一个字符串了，就可以归为一类了，代码如下所示。
+
+```java
+class Solution {
+    public List<List<String>> groupAnagrams(String[] strs) {
+        if (strs == null || strs.length == 0) return Collections.emptyList();
+        List<List<String>> list = new ArrayList<>();
+        Map<String, Integer> hash = new HashMap<>();
+        int i = 0;
+        for (String str : strs) {
+            char[] c = str.toCharArray();
+            Arrays.sort(c);
+            String sortStr = String.valueOf(c);
+            if (!hash.containsKey(sortStr)) {
+                hash.put(sortStr, i++);
+                List<String> sub = new ArrayList<>();
+                sub.add(str);
+                list.add(sub);
+            } else {
+                list.get(hash.get(sortStr)).add(str);
+            }
+        }
+        return list;
+    }
+}
+```
+
+
+## 结语
+
+如果你同我一样热爱数据结构、算法、LeetCode，可以关注我GitHub上的LeetCode题解：[awesome-java-leetcode][ajl]
+
+
+
+[title]: https://leetcode.com/problems/group-anagrams
+[ajl]: https://github.com/Blankj/awesome-java-leetcode

src/com/blankj/medium/_049/Solution.java

@@ -0,0 +1,44 @@
+package com.blankj.medium._049;
+
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.Collections;
+import java.util.HashMap;
+import java.util.List;
+import java.util.Map;
+
+/**
+ * <pre>
+ *     author: Blankj
+ *     blog  : http://blankj.com
+ *     time  : 2017/10/18
+ *     desc  :
+ * </pre>
+ */
+public class Solution {
+    public List<List<String>> groupAnagrams(String[] strs) {
+        if (strs == null || strs.length == 0) return Collections.emptyList();
+        List<List<String>> list = new ArrayList<>();
+        Map<String, Integer> hash = new HashMap<>();
+        int i = 0;
+        for (String str : strs) {
+            char[] c = str.toCharArray();
+            Arrays.sort(c);
+            String sortStr = String.valueOf(c);
+            if (!hash.containsKey(sortStr)) {
+                hash.put(sortStr, i++);
+                List<String> sub = new ArrayList<>();
+                sub.add(str);
+                list.add(sub);
+            } else {
+                list.get(hash.get(sortStr)).add(str);
+            }
+        }
+        return list;
+    }
+
+    public static void main(String[] args) {
+        Solution solution = new Solution();
+        System.out.println(solution.groupAnagrams(new String[]{"eat", "tea", "tan", "ate", "nat", "bat"}));
+    }
+}