2014-03-21 20:49
题目:设计一种排序算法,使得anagram排在一起。
解法:自定义一个comparator,使用额外的空间来统计字母个数,然后比较字母个数。
代码:
1 // 11.2 Sort an array of strings such that anagrams stay next to each other. 2 #include <algorithm> 3 #include <iostream> 4 #include <string> 5 #include <vector> 6 using namespace std; 7 8 string ta, tb; 9 int counting[256]; 10 11 void countingSort(string &s) 12 { 13 int i, j; 14 15 for (i = 0; i < 256; ++i) { 16 counting[i] = 0; 17 } 18 for (i = 0; i < (int)s.length(); ++i) { 19 ++counting[s[i]]; 20 } 21 s.clear(); 22 23 for (i = 0; i < 256; ++i) { 24 for (j = 0; j < counting[i]; ++j) { 25 s.push_back(i); 26 } 27 } 28 } 29 30 bool anagramComparator(const string &a, const string &b) 31 { 32 ta = a; 33 tb = b; 34 sort(ta.begin(), ta.end()); 35 sort(tb.begin(), tb.end()); 36 37 return ta < tb; 38 } 39 40 int main() 41 { 42 vector<string> v; 43 int i; 44 int n; 45 46 while (cin >> n && n > 0) { 47 v.resize(n); 48 for (i = 0; i < n; ++i) { 49 cin >> v[i]; 50 } 51 sort(v.begin(), v.end(), anagramComparator); 52 for (i = 0; i < n; ++i) { 53 if (i > 0) { 54 cout << ' '; 55 } 56 cout << v[i]; 57 } 58 cout << endl; 59 } 60 61 return 0; 62 }