LeetCode - Anagrams

转载于 2013-12-15 22:19:00 发布 · 68 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/zhuli19901106/p/3475842.html

本文介绍了一种通过排序字符串并按字母顺序分组的方法来识别一组字符串中的字母组合。使用了C++实现，并讨论了时间复杂性和空间复杂度。

Anagrams

2013.12.15 22:00

Given an array of strings, return all groups of strings that are anagrams.

Note: All inputs will be in lower-case.

Solution:

　　Anagrams means that strings composed of the same group of letters, such as [ate, eat, tea]. They'll become the same when sorted: aet.

　　My solution is to sort every string, and sort the array according to the sorted form of each string element. Then group the strings together by anagrams.

　　Time compexity is O(n * log(n)), where n is the length of the array. Space complexity is O(n).

Accepted code:

 1 // 1CE, 1AC
 2 #include <algorithm>
 3 #include <cstring>
 4 #include <string>
 5 using namespace std;
 6 
 7 typedef struct st{
 8 public:
 9     string str;
10     string sorted_str;
11     
12     st(string _str = "", string _sorted_str = "") {
13         str = _str;
14         sorted_str = _sorted_str;
15     }
16 }st;
17 
18 bool comparator(const st &x, const st &y)
19 {
20     // 1CE here, return a.sorted_str < b.sorted_str;
21     return x.sorted_str < y.sorted_str;
22 }
23 
24 class Solution {
25 public:
26     vector<string> anagrams(vector<string> &strs) {
27         // IMPORTANT: Please reset any member data you declared, as
28         // the same Solution instance will be reused for each test case.
29         int max_len = 0;
30         int i, j, n;
31         char *ps = nullptr;
32         
33         result.clear();
34         v.clear();
35         
36         n = strs.size();
37         if(n <= 0){
38             return result;
39         }
40         for(i = 0; i < n; ++i){
41             if(strs[i].length() > max_len){
42                 max_len = strs[i].length();
43             }
44         }
45         ps = new char[max_len + 1];
46         
47         string s, ss;
48         for(i = 0; i < n; ++i){
49             s = strs[i];
50             strcpy(ps, s.data());
51             sort(ps, ps + s.length());
52             ss = string(ps);
53             v.push_back(st(s, ss));
54         }
55         sort(v.begin(), v.end(), comparator);
56         delete[] ps;
57         i = 0;
58         while(i < n){
59             j = i;
60             while(j < n && v[i].sorted_str == v[j].sorted_str){
61                 ++j;
62             }
63             if(j - i > 1){
64                 while(i < j){
65                     result.push_back(v[i].str);
66                     ++i;
67                 }
68             }
69             i = j;
70         }
71         v.clear();
72         
73         return result;
74     }
75 private:
76     vector<st> v;
77     vector<string> result;
78 };