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