[LeetCode] 249. Group Shifted Strings 分组偏移字符串
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"]
,
Return:
[
["abc","bcd","xyz"],
["az","ba"],
["acef"],
["a","z"]
]
Note: For the return value, each inner list's elements must follow the lexicographic order.
一个字符串可以通过偏移变成另一个字符串,比如 ‘abc’ –> ‘bcd’ (所有字母右移一位),把可通过偏移转换的字符串归为一组。给定一个 String 数组,返回分组结果。
解法:将每个字符串都转换成与字符串首字符ASCII码值差的字符串,比如:'abc'就转换成'012','bcd'转换成了'012',两个就是同组的偏移字符串。用Hashmap来统计,key就是转换后的数字字符串,value是所有可以转换成此key的字符串集合。
注意:这个差值可能是负的,说明后面的字符比前面的小,此时加上26。
Java:
class Solution {
public List<List<String>> groupStrings(String[] strings) {
List<List<String>> result = new ArrayList<List<String>>();
HashMap<String, ArrayList<String>> map
= new HashMap<String, ArrayList<String>>(); for(String s: strings){
char[] arr = s.toCharArray();
if(arr.length>0){
int diff = arr[0]-'a';
for(int i=0; i<arr.length; i++){
if(arr[i]-diff<'a'){
arr[i] = (char) (arr[i]-diff+26);
}else{
arr[i] = (char) (arr[i]-diff);
} }
} String ns = new String(arr);
if(map.containsKey(ns)){
map.get(ns).add(s);
}else{
ArrayList<String> al = new ArrayList<String>();
al.add(s);
map.put(ns, al);
}
} for(Map.Entry<String, ArrayList<String>> entry: map.entrySet()){
Collections.sort(entry.getValue());
} result.addAll(map.values()); return result;
}
}
Python: Time: O(nlogn), Space: O(n)
import collections class Solution:
# @param {string[]} strings
# @return {string[][]}
def groupStrings(self, strings):
groups = collections.defaultdict(list)
for s in strings: # Grouping.
groups[self.hashStr(s)].append(s) result = []
for key, val in groups.iteritems():
result.append(sorted(val)) return result def hashStr(self, s):
base = ord(s[0])
hashcode = ""
for i in xrange(len(s)):
if ord(s[i]) - base >= 0:
hashcode += unichr(ord('a') + ord(s[i]) - base)
else:
hashcode += unichr(ord('a') + ord(s[i]) - base + 26)
return hashcode
C++:
class Solution {
public:
vector<vector<string>> groupStrings(vector<string>& strings) {
vector<vector<string> > res;
unordered_map<string, multiset<string>> m;
for (auto a : strings) {
string t = "";
for (char c : a) {
t += to_string((c + 26 - a[0]) % 26) + ",";
}
m[t].insert(a);
}
for (auto it = m.begin(); it != m.end(); ++it) {
res.push_back(vector<string>(it->second.begin(), it->second.end()));
}
return res;
}
};
类似题目:
[LeetCode] 49. Group Anagrams 分组变位词
[LeetCode] 300. Longest Increasing Subsequence 最长递增子序列
All LeetCode Questions List 题目汇总
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