catalogue
Interview question 01.01. Determine whether the character is unique
Sword finger Offer 50. The first character that appears only once
Interview question 01.02. Determine whether it is character rearrangement
1941. Check that all characters appear the same number of times
Sword finger Offer II 032. Effective modifier
1832. Judge whether the sentence is a full letter sentence
2053. The K-th unique string in the array
Zero, write in front
This is the 24th day of clocking in. There are many topics today. The main knowledge points are
1, Main knowledge points
1. Character counting
The hash table can be used to record the number of strings. If the string contains only lowercase letters, it can be further compressed.
int hash[26] ={0}; //hash table incidental initialization
for(int i = 0;s[i];++i)
hash[s[i] - 'a']++; //Statistical quantity2, After class exercises
Interview question 01.01. Determine whether the character is unique
Using the hash table, if the scanned letters are scanned, it will directly return false, and if it is not encountered after all scanning, it will return true.
bool isUnique(char* astr){
bool hash[256];
memset(hash,0,sizeof(hash));
for(int i = 0;astr[i];++i)
if(!hash[astr[i]]) hash[astr[i]]++;
else return false;
return true;
}Result analysis
Make do with
Sword finger Offer 50. The first character that appears only once
First use the hash table to count the occurrence times of all letters, and then scan the hash table. If there is one, it will be returned
Returns a space if there is no.
This question returns a character! So you can directly return a value!
char firstUniqChar(char* s){
int hash[26] ={0}; //hash table + initialization
for(int i = 0;s[i];++i) //Record hash value
hash[s[i] - 'a']++;
for(int i = 0;s[i];++i) //Traversal hash table
if(hash[s[i] - 'a'] == 1) return s[i];
return ' '; //Return space not found
}Result analysis
It's past
383. Ransom letter
383. Ransom letter
https://leetcode-cn.com/problems/ransom-note/ thinking
This time, it's just a different way to update the hash table
First, scan the magazine and insert all the elements and numbers into the hash table
Then, scan ransom to one character and subtract the corresponding hash table value by 1
bool canConstruct(char * ransomNote, char * magazine){
int hash[256] = {0}; //hash table
for(int i = 0;magazine[i];++i)//Scan magazine update hash table
hash[magazine[i]] ++;
for(int i = 0;ransomNote[i];++i) //Scan ransom update hash table
if(!hash[ransomNote[i]]) return false;//Not enough corresponding elements?
else hash[ransomNote[i]]--;
return true;
}Result analysis
It's OK
771. Gemstones and stones
771. Gemstones and stoneshttps://leetcode-cn.com/problems/jewels-and-stones/ thinking
First scan all elements in J, and then judge the elements in s according to the hash table for statistics.
int numJewelsInStones(char * jewels, char * stones){
bool has[256] = {0}; //hash table
int ans = 0;
for(int i = 0; jewels[i]; ++i) //Update hash table
if(has[jewels[i]] == 0) has[jewels[i]] = 1;
for(int i = 0; stones[i]; ++i)//According to hash statistics
if(has[stones[i]] == 1) ans++;
return ans;
}Result analysis
fairish
Interview question 01.02. Determine whether it is character rearrangement
If the number of corresponding elements is the same and the length of two strings is the same, the condition is satisfied.
bool CheckPermutation(char* s1, char* s2){
int hash[256] = {0},count = 0;//hash table
for(int i = 0;s1[i];++i){ //View all elements of s1
hash[s1[i]]++;
count++;
}
for(int i = 0;s2[i];++i){ //Compare whether the elements in s2 and s1 are the same
if(hash[s2[i]]) hash[s2[i]]--;//Found return
else return false;
if(!(count--)) return false;
}
if(count == 0)return true;//Equal length
else return false; //s1 is longer than s2
}Result analysis
It's OK
1941. Check that all characters appear the same number of times
First save the occurrence times of all elements in the hash table, and then check whether the non-0 elements in the hash table are equal to the maximum value.
bool areOccurrencesEqual(char * s){
int max = 0,hash[26] = {0};//Initialization of hash table and maximum value
for(int i = 0;s[i];++i){ //statistical information
hash[s[i]-'a']++;
if(hash[s[i]-'a'] > max) max = hash[s[i]-'a'];
}
for(int i = 0;i < 26;i++) //See if the requirements are met
if(hash[i] && hash[i] != max) return false;
return true;
}Result analysis
Make do with
242. Valid acronyms
242. Valid acronymshttps://leetcode-cn.com/problems/valid-anagram/ thinking
Go in two steps.
First determine whether all elements have
Then judge whether the length is the same
bool isAnagram(char * s, char * t){
int hash[26] = {0};
for(int i = 0;s[i];++i) //Insert hash statistics
hash[s[i] - 'a']++;
for(int i = 0;t[i];++i) //See if all elements are enough
if(hash[t[i] - 'a']) hash[t[i]-'a']--;
else return false;
for(int i = 0;i < 26;++i)
if(hash[i]) return false;//The hash table also has values indicating different lengths
return true;
}Result analysis
Make do with
Sword finger Offer II 032. Effective modifier
Sword finger Offer II 032. Effective modifierhttps://leetcode-cn.com/problems/dKk3P7/ thinking
Go in two steps
Count whether the number of occurrences of each character is the same
Determine whether the length is equal or identical
bool isAnagram(char * s, char * t){
int hash[26] = {0};
int ssize= 0,tsize = 0;
for(ssize = 0;s[ssize];++ssize)//Statistical words
hash[s[ssize] - 'a']++;
for(tsize = 0;t[tsize];++tsize)//See if the words appear the same number of times
if(hash[t[tsize] - 'a']) hash[t[tsize]-'a']--;
else return false;
if(ssize != tsize || !strcmp(s,t)) return false; //Are the lengths the same and not exactly the same
return true;
}Problem solving ideas
Make do with
1832. Judge whether the sentence is a full letter sentence
First, use the hash table to count all the letters that appear,
Just scan the hash table to see if there are any letters that don't appear?
bool checkIfPangram(char * sentence){
char hash[26] = {0};
for(int i = 0;sentence[i];++i)//View the elements that appear
hash[sentence[i] - 'a'] = true;
for(int i = 0;i < 26;i++) //See if there are elements that do not appear
if(!hash[i]) return false;
return true;
}Problem solving ideas
Starfish
2053. The K-th unique string in the array
Use a character pointer array to save different strings
When scanning the same character, the corresponding hashnum value will be + 1
Scan the entire hash table to find the hashnum value that only appears once, find a k-1, and finally return the corresponding hash value. If it cannot be found, an empty string of an application will be returned
char * kthDistinct(char ** arr, int arrSize, int k){
int count = 0,ansnum = 0;
char * hash[1000]; //Collect different strings
int hashnum[1000]= {0}; //Collect the number of occurrences of the corresponding string
for(int i = 0;i < arrSize;i++){ //Scan all ARRS for hash and hashnum insertion
int j = 0;
for(j = 0;j < count;j++)
if(strcmp(hash[j],arr[i]) == 0){//Appeared
hashnum[j] ++;
break;
}
if(j == count) { //Not present
hash[count++] = arr[i];
hashnum[count - 1] ++;
}
}
for(int i = 0,j = k;j > 0 && i < count;i++){ //Find the corresponding string address
if(hashnum[i] == 1)j--;
if(j == 0) return hash[i];
}
char *s = malloc(sizeof(char));//Empty string must be applied
s[0] = 0;
return s;
}Result analysis
Finished, finally!
Write at the end
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