题目描述 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&…

Jewels and Stones You're given strings J representing the types of stones that are jewels, and S representing the stones you have.  Each character inS is a type of stone you have.  You want to know how many of the stones you have are also jewels. The…

题目标签:Hash Table 这一题很简单,题目给了两个string:J 和 S. 只要把J 里面的 char 放入HashSet,再遍历S找出有多少个石头是宝石. Java Solution: Runtime beats 97.77% 完成日期:03/06/2019 关键点:HashSet class Solution { public int numJewelsInStones(String J, String S) { int result = 0; Set<Character> je…

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…

题目要求 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 …

Question 771. Jewels and Stones Solution 题目大意:两个字符串J和S,其中J中每个字符不同,求S中包含有J中字符的个数,重复的也算 思路:Set记录字符串J中的每个字符,遍历S中的字符,如果出现在Set中,count加1 Java实现: public int numJewelsInStones(String J, String S) { Set<Character> set = new HashSet<>(); int count = 0;…

题目描述: 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…

problem 771. Jewels and Stones solution1: class Solution { public: int numJewelsInStones(string J, string S) { ; unordered_set<char> js(J.begin(), J.end()); for(auto ch:S) { if(js.find(ch)!=js.end()) res++; } return res; } }; 参考 1. Leetcode_easy_771…

Jewels and Stones Description 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…

方法一:双指针 解题思路 假设链表存在相交时,headA 的长度为 a + c,headB 的长度为 b + c.如果把 headA 连上 headB,headB 连上 headB 的话,当遍历这两个新链表时有: \[(a + c) + (b + c) = (b + c) + (a + c) \] 而 \(a + c + b = b + c + a\),就出现相交的位置,因为 c 是相交部分的长度. 假设链表不相交,那么最后也会"相交","相交"于链表的尾部,即 n…