题目
Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.
Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.
The order of output does not matter.
Example 1:
Input:
s: "cbaebabacd" p: "abc"Output:
[0, 6]Explanation:
The substring with start index = 0 is "cba", which is an anagram of "abc".
The substring with start index = 6 is "bac", which is an anagram of "abc".
Example 2:
Input:
s: "abab" p: "ab"Output:
[0, 1, 2]Explanation:
The substring with start index = 0 is "ab", which is an anagram of "ab".
The substring with start index = 1 is "ba", which is an anagram of "ab".
The substring with start index = 2 is "ab", which is an anagram of "ab".
代码
#include "stdafx.h"
#include <list>
#include <string>
#include <map>
#include <iostream>
using namespace std;
class Solution {
public:
list<int> findAnagrams(string s, string t) {
list<int> result;
if (s.length() < t.length()) return result;
map<char, int> t_map;
for (int i = 0; i < t.length(); i++) {
if (t_map.find(t[i]) == t_map.end()) {
t_map[t[i]] = 1;
}
else {
t_map[t[i]] += 1;
}
}
int begin = 0, end = 0;
int len = INT_MAX;
while (begin < (s.length()-t.length()+1)) {
char c = s[begin];
if (t_map.find(c) != t_map.end()) {
int counter = t_map.size();
map<char, int> t_map_temp = t_map;
t_map_temp[c] --;
if (t_map_temp[c] == 0) counter--;
end = begin + 1;
while (counter > 0) {
char c_end = s[end];
if (t_map_temp.find(c_end) != t_map_temp.end()) {
t_map_temp[c_end] --;
if (t_map_temp[c_end] == 0) counter--;
if (t_map_temp[c_end] == -1) {
counter = -1;
}
}
else {
counter = -1;
}
end++;
}
if (counter == 0) {
result.push_back(begin);
}
}
begin++;
}
return result;
}
};
int main()
{
Solution solution;
list<int> result = solution.findAnagrams("euweiabbwoeiwopbabwepwoebbaase", "bba");
list<int>::iterator iter;
cout << result.size() << endl;
for (iter = result.begin(); iter != result.end(); iter++)
{
cout << *iter << endl;
}
}