How to solve the problem of gemstones and stones in leetcode
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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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