What I learned this week
- In vector, sort() sorting problem of strings (compared by alphabet in order, the first priority is higher: ABC < ACB < BCA)
- Sort problem in array: (sort function scope is [begin, end], excluding the last bit)
Static array: int a [10] = {9, 0, 1, 2, 3, 7, 4, 5, 8, 6}; sort (a, a + 10);
(note that length is added to the next bit at the end of the array)
vector: sort(heights.begin(), heights.end());
(end() in vector represents the next digit of the last digit) - string.substr (int start_ Usage of index, int length):
When there are two parameters, the first one is the starting position, and the second one is the substring length eg: substr(0,5) gets the string with the length of 5 starting from the 0 bit in string s
When a parameter is used, the string from the first bit to the end in string s is obtained by default from the specified start position to the end eg: substr(1)) - char string array ends with '\ 0', but string is a class with no character '\ 0' at the end
- Several methods to delete the last character of string in C + +:
- str.pop_back();
- str.erase(str.end() - 1);
- str = str.substr(0, str.length() - 1)； - string.erase() usage
(1) erase(pos,n); delete n characters from pos, for example, erase(0,1) is to delete the first character
(2) erase(first, last); delete characters from first to last (first and last are iterators)
(3) erase(pos); delete a character at pos (pos must be an iterator of type string string::iterator pos) - The algorithm for finding prime numbers: eradorse sieve (a simplified and efficient algorithm for finding multiples of each number)
- On character transcoding:
Char to int: generally use char - '0'
Int to cha: generally use int + '0'
(Note: char(4): convert 4 to '\ 4' instead of character '4', int('a '): convert string to corresponding ASCII code
Therefore, when it comes to int/char conversion, it is better to use ± '0', where (char) should never be used - Judge whether there is a value in vector: (STL) find
vector<int>::iterator it = find(arr.begin(),arr.end(),res);
if( it != arr.end()) / / value in array
if( it == arr.end()) / / value not in array - Freud's circular search algorithm (fast and slow pointer method / tortoise and rabbit pointer method)
Principle: every step the rabbit takes, it approaches the tortoise to a node (in the direction of their movement). If there is, we will meet
Used to determine whether a linked list has cycles
594, longest harmonic subsequence
class Solution { public: int findLHS(vector<int>& nums) { unordered_map<int, int> hash; for(auto num:nums) hash[num]++; int res = 0; for(auto i : hash) //if(hash[i.first+1] != 0) if(hash.count(i.first+1)) res = max(res, i.second + hash[i.first + 1]); return res; } };
1160, spelling words
class Solution { public: int countCharacters(vector<string>& words, string chars) { int res=0; int ch[26]={0}; for(char c:chars) ch[c-'a']++; for(string word:words) { int wd[26]={0}; for(char c:word) wd[c-'a']++; res+=word.size(); for(char c:word) { if(ch[c-'a']<wd[c-'a']) { res-=word.size(); break; } } } return res; } };
572, candy
class Solution { public: int distributeCandies(vector<int>& candies) { int nums = candies.size(); vector<int> hash(10000); for(int i = 0; i < nums; i++){ hash[candies[i]]++; } int res = 0; for(int i = 0; i < 10000; i++){ if(hash[i] >= 2) res++; else if(hash[i] == 1 && res < (nums/2) ) res++; } return min(nums/2, res); } };
599 minimum index sum of common elements of two lists
class Solution { public: vector<string> findRestaurant(vector<string>& list1, vector<string>& list2) { unordered_map<string, int> hash; //unordered_map<string, unordered_map> hash; int i = 0; for(auto r1 : list1){ i++; hash[r1] += i; } int j = 0; for(auto r2 : list2){ j++; if(hash[r2] != 0){ int tmp = -hash[r2]; hash[r2] = tmp; hash[r2] -= j; } } int min_res = INT_MAX; for(auto h:hash){ if(h.second < 0) min_res = min(min_res, -h.second); } vector<string> arr_res; for(auto h:hash){ if(-h.second == min_res) arr_res.push_back(h.first); } return arr_res; } };
1078. Bigram participle
There are many small details about this question:
- char string array ends with '\ 0', but string is a class, and there is no character '\ 0' at the end (string ends with string length), so a '\ 0' is added to the string of this question as the end identifier
- Common operations in string:
Delete: ear (iterator, Iterator);
Add the character append (number, char) at the end;
class Solution { public: vector<string> findOcurrences(string text, string first, string second) { vector<string> str_arr; string str; text.append(1,'\0'); for(auto c : text){ if(c == ' ' || c == '\0'){ str_arr.push_back(str); str.erase (str.begin(), str.end()); } else str.append(1, c); } vector<string> res; for(int i = 1; i<str_arr.size(); i++){ if(str_arr[i - 1] == first && str_arr[i] == second && i != str_arr.size() - 1) res.push_back(str_arr[i + 1]); } return res; } };
645. Wrong set
Note: the initialization of hash table and the specified operation are carried out separately to achieve the purpose
class Solution { public: vector<int> findErrorNums(vector<int>& nums) { unordered_map<int, int> hash; for(int i = 0; i < nums.size();i++){ hash[i] = 0; } for(auto num:nums){ hash[num-1]++; } vector<int> res; for(auto h:hash){ if(h.second == 2) res.push_back(h.first+1); } for(auto h:hash){ if(h.second == 0) res.push_back(h.first+1); } return res; } };
720, the longest word in the dictionary
const int MAX_NODE = 1000000 + 10; const int CHARSET = 26; int trie[MAX_NODE][CHARSET] = {0}; int ending[MAX_NODE] = {0}; int k = 1; // k is the node mark void insert(string w){ //Insert a word in the Trie tree int len = w.size(); int p = 0; // p is the dynamic node tag for(int i=0; i<len; i++){ int c = w[i] - 'a'; if(!trie[p][c]){ trie[p][c] = k; k++; } p = trie[p][c]; } ending[p] = 1; //The ending array is a hash table, which is recorded as the node mark k of the terminating node } int search(string s){ int len = s.size(); int p = 0; for(int i=0; i<len; i++){ int c = s[i] - 'a'; if(!trie[p][c]) return 0; p = trie[p][c]; } return ending[p] == 1; } class Solution { public: string longestWord(vector<string>& words) { for(auto word:words){ insert(word); } vector<string> res_arr; for(auto word:words){ int length = word.size(); bool flag = 1; while(length){ if(!search(word.substr(0,length--))){ flag = 0; break; } } if(flag) res_arr.push_back(word); } int max_len = 0; for(auto s:res_arr){ int s_len = s.size(); max_len = max(max_len, s_len); } vector<string> res; for(auto s:res_arr){ if(s.size() == max_len) res.push_back(s); } sort(res.begin(), res.end()); return res[0]; } };
204 finding prime numbers
Eradosse method: efficient for loop to create hash table
class Solution { public: int countPrimes(int n) { int count = 0; //Initial default all numbers are prime vector<bool> signs(n, true); for (int i = 2; i < n; i++) { if (signs[i]) { count++; for (int j = i + i; j < n; j += i) { //Exclude numbers that are not prime signs[j] = false; } } } return count; } };
202 happy number
Happiness number: the sum of squares of each digit is 1, non-1 and can continue to cycle, and finally 1)
tips: in order to prevent infinite loop, you can use the array to memorize the values that have appeared, and reappear to indicate the dead loop
class Solution { public: bool isHappy(int n) { unordered_map<char, int> hash; for(int i =0; i<10;i++){ hash[i+'0'] = i*i; //hash[char(i)] error -- (char) converts 4 to '\ 4' instead of '4'; } int tacket = n; vector<int> arr; while(1){ string num = to_string(tacket); int res = 0; for(auto c:num){ res += hash[c]; } if(res == 1) return 1; vector<int>::iterator it=find(arr.begin(),arr.end(),res); if(it != arr.end()) return 0; else arr.push_back(res); tacket = res; } return 0; } };