leetcode 204/187/205 Count Primes/Repeated DNA Sequences/Isomorphic Strings

今天药忘吃喽~ 2022-08-08 05:27 165阅读 0赞

一：leetcode 204 [Count Primes][]

题目：

Description:

Count the number of prime numbers less than a non-negative number, n

分析：此题的算法源码可以参看这里，http://en.wikipedia.org/wiki/Sieve\_of\_Eratosthenes

代码：

class Solution {
    public:
        int countPrimes(int n) {     // 求小于一个数n的素数个数 方法思想
            bool *isPrimes = new bool[n];
            memset(isPrimes, 0, sizeof(bool)*(n));
            int ans = 0;
            for(int i = 2; i < n; i++){
                while(i < n && isPrimes[i]) i++;
    			if(i >= n) break;
                ans ++;
                for(int j = 2; j <= (n-1)/i; j++)
                    isPrimes[j*i] = 1;
            }
            delete []isPrimes;
            return ans;
        }
    };

二：leetcode 187 [Repeated DNA Sequences][]

题目：

All DNA is composed of a series of nucleotides abbreviated as A, C, G, and T, for example: "ACGAATTCCG". When studying DNA, it is sometimes useful to identify repeated sequences within the DNA.

Write a function to find all the 10-letter-long sequences (substrings) that occur more than once in a DNA molecule.

For example,

Given s = "AAAAACCCCCAAAAACCCCCCAAAAAGGGTTT",
    
    Return:
    ["AAAAACCCCC", "CCCCCAAAAA"].

分析：此题可以看做是判重，可以采用hash\_table或者trie树，但是此题对内存要求比较严格，用map<string, int>则会memory limit，此外我这里写的trie树也会memory limit,

因此使用hashtable的话就必须找到hash函数将string转换为int，一是可以自己编写hash函数或者是用c++11自带的hash<string>。

代码：

一：hashtable

class Solution {
    public:
        vector<string> findRepeatedDnaSequences(string s) {
            vector<string> result;
            hash<string> hash_func;            // 将string转换为int的hash函数
            for(int i = 0; i < s.size(); i++){
                string str = s.substr(i, 10);
                if(str.size() < 10)
                    break;
               int x = hash_func(str);
    			if(hmap.count(x)){
    				if(hmap[x] == 1)result.push_back(str);
    				hmap[x]++;
    			} 
                else hmap.insert(pair<int, int>(x, 1));
            }
            return result;
        }
    private:
        unordered_map<int, int> hmap;  // c++的STL中map/set用string会消耗很多memory
    };

或者用自己写的hash table：

class Solution {
    public:
    	int strToInt(const string &str){
    		int result = 0;
    		for(int i = 0; i < 10; i++){
    			switch(str[i])
    			{
    			case 'A':
    				result += 0;
    				break;
    			case 'C':
    				result += 1;
    				break;
    			case 'G':
    				result += 2;
    				break;
    			case 'T':
    				result += 3;
    				break;
    			}
    			result = (result << 2);
    		}
    		return result;
    	}
        vector<string> findRepeatedDnaSequences(string s) {
            vector<string> result;
            hash<string> hash_func;            // 将string转换为int的hash函数
            for(int i = 0; i < s.size(); i++){
                string str = s.substr(i, 10);
                if(str.size() < 10)
                    break;
               // int x = hash_func(str);
    			int x = strToInt(str);
    			if(hmap.count(x)){
    				if(hmap[x] == 1)result.push_back(str);
    				hmap[x]++;
    			} 
                else hmap.insert(pair<int, int>(x, 1));
            }
            return result;
        }
    private:
        unordered_map<int, int> hmap;  // c++的STL中map/set用string会消耗很多memory
    };

二：trie树的代码——会memory limit

struct TrieNode{
        TrieNode *node[5];
        int count;
        TrieNode():count(1){
            for(int i = 0; i < 5; i++){
                node[i] = NULL;
            }
        }
    };
    
    
    class Solution {
    private:
        void destroy(TrieNode *p){
            if(p != NULL){
                for(int i = 0; i < 5; i++)
    				destroy(p->node[i]);
                delete p;
            }
        }    
    public:
        bool isExist(TrieNode *p, const string &str){
            bool flag = true;
            for(int i = 0; i < str.size(); i++){
                int index = (str[i]-'A') % 5;
                if(p->node[index] == NULL){
                    p->node[index] = new TrieNode();
                    flag = false;
                }
                p = p->node[index];
            }
            if(flag){
                if(p->count != 1) flag = false;
                p->count ++;
            }
            return flag;
            
        }
        vector<string> findRepeatedDnaSequences(string s) {
            vector<string> result;
            TrieNode *root = new TrieNode();
            for(int i = 0; i < s.size(); i++){
                string str = s.substr(i, 10);
                if(str.size() < 10) break;
                if(isExist(root, str)) result.push_back(str);
            }
            destroy(root);
            return result;
        }
        
    };

三：leetcode 205 [Isomorphic Strings][]

题目：

Given two strings s and t, determine if they are isomorphic.

Two strings are isomorphic if the characters in s can be replaced to get t.

All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character but a character may map to itself.

For example,  
Given `"egg"`, `"add"`, return true.

Given `"foo"`, `"bar"`, return false.

Given `"paper"`, `"title"`, return true.

Note:  
You may assume both s and t have the same length.

分析：此时是判断两个字符串是否同构，两边必须都是一一映射，所以我这里采用了两个hash\_map

class Solution {
    public:
        bool isIsomorphic(string s, string t) {
            if(s == t) return true;
            int m = s.size(), n = t.size();
            if(m != n) return false;
            for(int i = 0; i < m; i++){
                if(lmap.count(s[i]) && lmap[s[i]] != t[i]) return false;
                else
                   lmap.insert(pair<char, char>(s[i], t[i])); 
                   
                if(rmap.count(t[i]) && rmap[t[i]] != s[i]) return false;
                else
                   rmap.insert(pair<char, char>(t[i], s[i])); 
            }
            return true;
        }
    private:
        unordered_map<char, char> lmap;    // 分别建立了左右两个hash table
        unordered_map<char, char> rmap;
    };

[Count Primes]: https://leetcode.com/problems/count-primes/
[Repeated DNA Sequences]: https://leetcode.com/problems/repeated-dna-sequences/
[Isomorphic Strings]: https://leetcode.com/problems/isomorphic-strings/