The given string J represents the type of gem in the stone, and the string S represents the stone you own. Each character in S represents a type of stone you own, and you want to know how many of the stones you own are gems. The letters in J are not repeated, and all the characters in J and S are letters. Letters are case-sensitive, so "a" and "A" are different types of stones.
Example 1:
Input: J = "aA", S = "aAAbbbb"
Output: 3
Example 2:
Input: J = "z", S = "ZZ"
Output: 0
note:
S and J can contain up to 50 letters.
The characters in J are not repeated.
Source: LeetCode 771. Gems and Stones
Link: https://leetcode-cn.com/problems/jewels-and-stones/
Algorithm 1: Double for loop
int Fun(string J, string S) {
int i = 0;
for (int j = 0; j < J.length(); j++)
{
for (int s = 0; s < S.length(); s++)
{
if (J[j] == S[s])
i++;
}
}
return i;
}
Algorithm 2: c++ find() function
int numJewelsInStones(string J, string S) {
int cnt = 0;
for (auto s : S)
{
if (find(J.begin(), J.end(), s) != J.end())
{
cnt++;
}
}
return cnt;
}
Algorithm 3: Implement a hash (hash) table, use the hash table to record the types of gems, and traverse again to count the number of gems
int numJewelsInStones2(char * J, char * S)
{
int num = 0;//宝石数量
char chars[128] = {
0};//统计J中宝石是否出现,0没有出现,1出现 .hash表
//遍历J,记录宝石出现的种类
while(*J != '\0')
{
chars[*J++] = 1;
}
// 遍历石头，统计宝石的个数
while(*S != '\0')
{
if(chars[*S++] == 1)
{
num++;
}
}
return num;
}
Algorithm 4: Use count_if and any_if() functions.
Usage: Usage: bool any_of (InputIterator _First, InputIterator _Last, Predicate _Pred );
int numJewelsInStones(string J, string S) {
return count_if(S.begin(), S.end(),
[&](char ch) {
return any_of(J.begin(), J.end(), [&](char a) {
return a == ch; }); });
}