|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|

[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

Given an array of strings, group anagrams together.

For example, given:`["eat", "tea", "tan", "ate", "nat", "bat"]`,

Return:

**Note:** All inputs will be in lower-case.

**Tags:** Hash Table, String

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。

classSolution {

publicList<List<String>>groupAnagrams(String[]strs) {

if (strs==null|| strs.length==0)returnCollections.emptyList();

List<List<String>> list=newArrayList<>();

Map<String,Integer> hash=newHashMap<>();

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=newArrayList<>();

sub.add(str);

list.add(sub);

}else {

list.get(hash.get(sortStr)).add(str);

}

}

return list;

}

}

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

packagecom.blankj.medium._049;

importjava.util.ArrayList;

importjava.util.Arrays;

importjava.util.Collections;

importjava.util.HashMap;

importjava.util.List;

importjava.util.Map;

publicclassSolution {

publicList<List<String>>groupAnagrams(String[]strs) {

if (strs ==null ||strs.length ==0)returnCollections.emptyList();

List<List<String>>list =newArrayList<>();

Map<String,Integer>hash =newHashMap<>();

inti =0;

for (Stringstr :strs) {

char[]c =str.toCharArray();

Arrays.sort(c);

StringsortStr =String.valueOf(c);

if (!hash.containsKey(sortStr)) {

hash.put(sortStr,i++);

List<String>sub =newArrayList<>();

sub.add(str);

list.add(sub);

}else {

list.get(hash.get(sortStr)).add(str);

}

}

returnlist;

}

publicstaticvoidmain(String[]args) {

Solutionsolution =newSolution();

System.out.println(solution.groupAnagrams(newString[]{"eat","tea","tan","ate","nat","bat"}));

}

}