LeetCode 249. Group Shifted Strings （群组移位字符串）$

Given a string, we can "shift" each of its letter to its successive letter, for example: "abc" -> "bcd". We can keep "shifting" which forms the sequence:

"abc" -> "bcd" -> ... -> "xyz"

Given a list of strings which contains only lowercase alphabets, group all strings that belong to the same shifting sequence.

For example, given: ["abc", "bcd", "acef", "xyz", "az", "ba", "a", "z"], 
A solution is:

[
  ["abc","bcd","xyz"],
  ["az","ba"],
  ["acef"],
  ["a","z"]
]

---

题目标签：Hash Table

　　题目给了我们一个 strings array，让我们把不同移位距离的string 分开归类。

　　首先来看一下相同移位距离string 的特性：

　　　　相同的移位string，拥有相同的移位距离，比如abc, bcd, xyz 都是移位了1个距离。根据这个特性，我们可以把bcd 和 xyz 恢复到 abc。

　　利用HashMap，把最原始的 归位string 当作key，把可以恢复到 原始的 归位string 的 所有strings（List）当作value 存入map。

　　

Java Solution:

Runtime beats 44.74% 

完成日期：11/04/2017

关键词：HashMap

关键点：利用 char - 'a' 把所有相同移位距离的strings 转换成 同一个原始string 存入map

class Solution 
{
    public List<List<String>> groupStrings(String[] strings) 
    {
        List<List<String>> res = new ArrayList<>();
        
        HashMap<String, List<String>> map = new HashMap<>();
        
        // store original string as key; (List) strings come from same original one as value
        for(String str: strings)
        {
            int offset = str.charAt(0) - 'a';
            String key = "";
            
            for(int i=0; i<str.length(); i++)
            {
                char c = (char) (str.charAt(i) - offset);
                if(c < 'a')
                    c += 26;
                
                key += c;
            }
             
            if(!map.containsKey(key))
                map.put(key, new ArrayList<String>());
             
            map.get(key).add(str);
            
        }
        
        // add each key's value into res
        for(String key: map.keySet())
        {
            res.add(map.get(key));
        }
        
        return res;
    }
}

参考资料：

https://discuss.leetcode.com/topic/20722/my-concise-java-solution

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