249. Group Shifted Strings

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;
    }
}

二刷:

二刷就好一些。之前写得比较乱也不容易懂,下面来分析二刷的想法。 

  1. 主要是利用和求anagram一样的方法,利用HashMap来保存能group到一起的单词
  2. 对于能group到一起的每个单词来说,他们都会有一个base case,每个单词都是从这个base case shift出去的, 比如 abc, bcd, cde一类的,我们可以把abc认定为base case,那么所有base case为abc的单词,我们就可以group到一起
  3. 了解了原理,接下来我们就可以操作实现了,首先我们有一个HashMap<String,List<String>>, 这里的key是base case String, 比如abc, ab, a一类的, 而value List<String>就是可以被放入结果集中的List
  4. 我们遍历给定的数组Strings, 对于每一个单词s,
    1. 我们求出它的base case string
      1. 这里另外写了一个球base case string的方法getBaseStr
      2. 我们假定所有的base case string均以'a'为第一个字母
      3. 求出第一个字母被shift的长度lenShifted =  s.charAt(0) - 'a',  并且建立一个StringBuilder用来保存结果
      4. 接下来对于string s中的每一个字母c,我们计算他在被移动lenShifted个长度之前的字母是什么
        1. 我们可以用c - 'a'求出c到'a'的距离, 再减去这个lenShifted,就得到了当前这个字母究竟被shift了多少个单位curShifted
        2. 之后处理一下特殊情况,当这个curShifted < 0的时候,我们要加上26,把它重新映射到26个小写字母里
        3. 最后求出c = char('a' + curShifted),这时更新后的c就是 shift之前的base char
        4. 我们把这个base char存入StringBuilder里,接着计算下一个字符。  (这里也可以使用stefan pochmann的写法大大简化代码)
        5. 最后返回sb.toString()就是我们要求的base case string
    2. 得到base case string把单词按照base case string存入到hashmap里
  5. 遍历完毕以后,我们还要对HashMap中的每个group进行排序
  6. 最后利用hashMap的结果集合values生成最终结果, 即 res = new ArrayList<List<String>>(map.values());

Java:

Time Complexity - O(nlogn), 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;
        }
        Map<String, List<String>> map = new HashMap<>();
        for (String s : strings) {
            String base = getBaseStr(s);
            if (!map.containsKey(base)) {
                map.put(base, new ArrayList<>());
            }
            map.get(base).add(s);
        }
        for (String s : map.keySet()) {
            Collections.sort(map.get(s));
        }
        res = new ArrayList<List<String>>(map.values());
        return res;
    }
    
    private String getBaseStr(String s) {
        if (s == null || s.length() == 0) {
            return s;
        }
        StringBuilder sb = new StringBuilder();
        int lenShifted = s.charAt(0) - 'a';
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            int curShifted = c - 'a' - lenShifted;
            if (curShifted < 0) {
                curShifted += 26;
            }
            c = (char)('a' + curShifted);
            sb.append(c);
        }
        return sb.toString();
    }
}

Reference:

https://leetcode.com/discuss/50358/my-concise-java-solution

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/50163/1-4-lines-ruby-and-python

原文地址:https://www.cnblogs.com/yrbbest/p/5009004.html