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This article mainly shows you "how to solve the problem of gems and stones in leetcode", the content is simple and easy to understand, and the organization is clear. I hope it can help you solve your doubts. Let Xiaobian lead you to study and learn this article "how to solve the problem of gems and stones in leetcode".

Title Link

https://leetcode-cn.com/problems/jewels-and-stones/

Title Description

Given the 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 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 contain up to 50 letters.

The characters in J are not repeated.

solution train of thought

Tag: string

J is traversed first, and the characters are stored in HashSet respectively, so that they can be searched later when traversing S.

Traversing S and comparing each character with those in HashSet, if any, the result is ans++, the traversal ends, returning ans

Time complexity: O(m+n), m is the length of J, n is the length of S

code

Java version

class Solution {

public int numJewelsInStones(String J, String S) {

Set set = new HashSet();

for(int i = 0; i

< J.length(); i++) { set.add(J.charAt(i)); } int ans = 0; for(int i = 0; i < S.length(); i++) { if(set.contains(S.charAt(i))){ ans++; } } return ans; } } JavaScript版本 /** * @param {string} J * @param {string} S * @return {number} */ var numJewelsInStones = function(J, S) { const set = new Set(); for(const s of J) { set.add(s); } let ans = 0; for(const s of S) { if(set.has(s)){ ans++; } } return ans; }; 画解

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