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LeetCode刷题:771. Jewels and Stones
原题链接:
You're given strings J
representing the types of stones that are jewels, and S
representing the stones you have. Each character in S
is a type of stone you have. You want to know how many of the stones you have are also jewels.
The letters in J
are guaranteed distinct, and all characters in J
and S
are letters. Letters are case sensitive, so "a"
is considered a different type of stone from "A"
.
Example 1:
Input: J = "aA", S = "aAAbbbb"Output: 3
Example 2:
Input: J = "z", S = "ZZ"Output: 0
Note:
S
and J
will consist of letters and have length at most 50.J
are distinct.给定字符串J
代表石头中宝石的类型,和字符串 S
代表你拥有的石头。 S
中每个字符代表了一种你拥有的石头的类型,你想知道你拥有的石头中有多少是宝石。
J
中的字母不重复,J
和 S
中的所有字符都是字母。字母区分大小写,因此"a"
和"A"
是不同类型的石头。
示例 1:
输入: J = "aA", S = "aAAbbbb"输出: 3
示例 2:
输入: J = "z", S = "ZZ"输出: 0
注意:
S
和 J
最多含有50个字母。J
中的字符不重复。算法设计
package com.bean.algorithm.basic;public class JewelsandStones { /* * Input: J = "aA", S = "aAAbbbb" * Output: 3 * */ public int numJewelsInStones(String J, String S) { char[] jewels = J.toCharArray(); char[] stones = S.toCharArray(); int found = 0; for(int i = 0; i < stones.length; i++) { for(int j = 0; j < jewels.length; j++) { if(jewels[j] == stones[i]) { found++; } } } return found; } public static void main(String[] args) { // TODO Auto-generated method stub JewelsandStones jewelsandstones=new JewelsandStones(); String J= "aA"; String S="aAAbbbb"; int result = jewelsandstones.numJewelsInStones(J, S); System.out.println("result = "+result); }}
程序运行结果:
result = 3
LeetCode上提交代码,Accepted。
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