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题目：
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.
链接： http://leetcode.com/problems/group-shifted-strings/
题解：
一道简单题，我又写得巨长...水平太有限了，唉。主要是数据结构选了一个Map<Integer, Map<String, List<String>>>，先以String长度为key，再以单词为key，List<String>为value。 其实可以不用这么麻烦， 就用Map<String，List<String>>就可以了。遍历keySet()的时候，只要长度不一样，continue就行了。时间复杂度还是一样的。 假设单词长度有限的情况下，还是O(n2)。 二刷再简化..........这句话说了几百遍。二刷再简化吧！
Time Complexity - O(n2)， Space Complexity - O(n)。
public class Solution { public List<List<String>> groupStrings(String[] strings) { List<List<String>> res = new ArrayList<>(); if(strings == null || strings.length == 0) return res; Arrays.sort(strings); Map<Integer, Map<String, List<String>>> map = new HashMap<>(); for(int i = 0; i < strings.length; i++) { String s = strings[i]; int len = s.length(); if(!map.containsKey(len)) { Map<String, List<String>> tmpMap = new HashMap<>(); tmpMap.put(s, new ArrayList<>(Arrays.asList(s))); map.put(len, tmpMap); } else { Map<String, List<String>> tmpMap = map.get(len); boolean hasSequence = false; outerloop: for(String t : tmpMap.keySet()) { for(int k = 1; k < len; k++) { int curDistance = (int)(s.charAt(k) - t.charAt(k)); int lastDistance = (int)(s.charAt(k - 1) - t.charAt(k - 1)); curDistance = curDistance >= 0 ? curDistance : curDistance + 26; lastDistance = lastDistance >= 0 ? lastDistance : lastDistance + 26; if(curDistance != lastDistance) continue outerloop; } tmpMap.get(t).add(s); hasSequence = true; break; } if(!hasSequence) { tmpMap.put(s, new ArrayList<>(Arrays.asList(s))); } } } for(int i : map.keySet()) { Map<String, List<String>> tmpMap = map.get(i); for(String s : tmpMap.keySet()) { res.add(map.get(i).get(s)); } } return res; } }
Reference:
https://leetcode.com/discuss/69783/concise-10-lines-java-solution-with-explanation
https://leetcode.com/discuss/67240/around-13-lines-code-in-java
https://leetcode.com/discuss/64979/simple-solution-in-java-with-detailed-explaination
https://leetcode.com/discuss/64751/cannot-pass-tests-seems-nothing-wrong-for-the-custom-testcase
https://leetcode.com/discuss/58003/java-solution-with-separate-shiftstr-function
https://leetcode.com/discuss/53166/4ms-c-solution
https://leetcode.com/discuss/52627/python-easy-to-understand-solution-with-comments
https://leetcode.com/discuss/50582/4ms-c-solution
https://leetcode.com/discuss/50557/4ms-easy-c-solution-with-explanations
https://leetcode.com/discuss/50416/1-4-lines-in-java
https://leetcode.com/discuss/50358/my-concise-java-solution
https://leetcode.com/discuss/50163/1-4-lines-ruby-and-python
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原文地址：http://www.cnblogs.com/yrbbest/p/5009004.html
