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438_FindAll_Anagram_inaString.cpp
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438_FindAll_Anagram_inaString.cpp
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/* https://leetcode.com/problems/find-all-anagrams-in-a-string/
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.
class Solution(object):
def findAnagrams(self, s, p):
"""
:type s: str
:type p: str
:rtype: List[int]
"""
*/
#include <iostream>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
using namespace std;
std::vector<int> findAnagrams(string s, string p);
bool compare_count(char x[], char y[]);
int main(){
std::vector<int> result;
string s1 = "cbaebabacd";
string p1 = "abc";
result = findAnagrams(s1, p1);
for(int i = 0; i < result.size(); i++)
cout << result[i] << " ";
cout << endl;
string s2 = "abab";
string p2 = "ab";
result = findAnagrams(s2, p2);
for(int i = 0; i < result.size(); i++)
cout << result[i] << " ";
}
std::vector<int>findAnagrams(string s, string p){
std::vector<int> result;
int n = s.size();
int target = p.size();
// size of ASCII Code 256 - to make sure we can cover all the characters
char count_s[256] = {0};
char count_p[256] = {0};
// count s of window size p and p
for(int i = 0; i < target; i++){
(count_s[s[i]])++;
(count_p[p[i]])++;
}
// -------
// --
// Current
// --
// --
// --
// --
// --
for(int i = target; i < n; i++){
if(compare_count(count_s, count_p)) // if two count arrays correspond
result.push_back(i - target); // add element
// slide to next window size
// add next char, remove the first character
count_s[s[i]]++;
count_s[s[i-target]]--;
}
// Need to compare n - target + 1 times, so the last comparison
if(compare_count(count_s, count_p)) // if two count arrays correspond
result.push_back(n - target); // add element
return result;
}
bool compare_count(char x[], char y[]){
for(int i = 0; i < 256; i++){
if(x[i] != y[i])
return false;
}
return true;
}