[LeetCode]--49. Group Anagrams

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.

anagrams：由颠倒字母顺序而构成的字

用一个map来存放，map的key存放字符串按字母表字符排序后的顺序，value存放原本的字符串。

public List<List<String>> groupAnagrams(String[] strs) {        Map<String, ArrayList<String>> map = new HashMap<String, ArrayList<String>>();        for (String str : strs) {            String sortedStr = sortStr(str);            if (map.containsKey(sortedStr)) {                map.get(sortedStr).add(str);            } else {                map.put(sortedStr, new ArrayList<String>());                map.get(sortedStr).add(str);            }        }        List<List<String>> res = new ArrayList<List<String>>(map.values());        for (List<String> list : res) {            Collections.sort(list);        }        return res;    }    private String sortStr(String str) {        char[] charArr = str.toCharArray();        Arrays.sort(charArr);        return new String(charArr);    }

我用的是这个判断是否相同的组成。这种思路也可以供大家参考。

public boolean isTheSameComponent(String a, String b) {        if ((a == null || a.length() == 0) && (b == null || b.length() == 0))            return true;        if (a.length() != b.length())            return false;        List<Character> list = new ArrayList<Character>();        for (int i = 0; i < a.length(); i++)            list.add(a.charAt(i));        for (int i = 0; i < b.length(); i++)            if (list.contains(b.charAt(i))) {                list.remove((Character) b.charAt(i));            }        return list.size() == 0;    }