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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"],
Return:
[ ["abc","bcd","xyz"], ["az","ba"], ["acef"], ["a","z"] ]
Note: For the return value, each inner list‘s elements must follow the lexicographic order.
关键点：group all strings that belong to the same shifting sequence， 所以找到属于相同移位序列的key 很重要。因此单独写了函数shiftStr,
buffer.append((c - ‘a‘ - dist + 26) % 26 + ‘a‘) 极容易错。将相同移位序列的Strings存入key 对应的value中，建立正确的HashMap很重要。
Java code:
public class Solution { public List<List<String>> groupStrings(String[] strings) { if (strings == null) { throw new IllegalArgumentException("strings is null"); } List<List<String>> result = new ArrayList<List<String>>(); if(strings.length == 0){ return result; } Map<String, List<String>> map = new HashMap<String, List<String>>(); for(String str: strings) { String shifted_str = shiftStr(str); if(map.containsKey(shifted_str)){ map.get(shifted_str).add(str); }else{ List<String> item = new ArrayList<String>(); item.add(str); map.put(shifted_str, item); result.add(item); } } for(List<String> list: result) { Collections.sort(list); } return result; } private String shiftStr(String str){ StringBuilder buffer = new StringBuilder(); char[] char_array = str.toCharArray(); int dist = str.charAt(0) - ‘a‘; for(char c: char_array){ buffer.append((c - ‘a‘ - dist + 26) % 26 + ‘a‘); } return buffer.toString(); } }
Reference:
1. https://leetcode.com/discuss/58003/java-solution-with-separate-shiftstr-function
Leetcode Group Shifted Strings
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原文地址：http://www.cnblogs.com/anne-vista/p/4868855.html
