(找了leetcode上最简单的一个题来找一下存在感) 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…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述: 题目大意 解题方法 数组count 字典Counter 日期 题目地址:https://leetcode.com/problems/jewels-and-stones/description/ 题目描述: You're given strings J representing the types of stones that are jewels, and S rep…

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…

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&…

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;…

[LeetCode]Minimum Depth of Binary Tree Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. 递归和非递归,此提比较简单.广度优先遍历即可.关键之处就在于如何保持访问深度. 下面是4种代码: im…

Given an index k, return the kth row of the Pascal's triangle. For example, given k = 3, Return [1,3,3,1]. Note: Could you optimize your algorithm to use only O(k) extra space? 思路:最简单的方法就是依照[Leetcode]Pascal's Triangle 的方式自顶向下依次求解,但会造成空间的浪费.若仅仅用一个vect…

53. Maximum Subarray[leetcode] Find the contiguous subarray within an array (containing at least one number) which has the largest sum. For example, given the array [-2,1,-3,4,-1,2,1,-5,4],the contiguous subarray [4,-1,2,1] has the largest sum = 6. 挑…

27. Remove Element[leetcode] Given an array and a value, remove all instances of that value in place and return the new length. Do not allocate extra space for another array, you must do this in place with constant memory. The order of elements can b…

[刷题][LeetCode]总 用动画的形式呈现解LeetCode题目的思路 参考链接-空 007-整数反转 方法: 弹出和推入数字 & 溢出前进行检查 思路: 我们可以一次构建反转整数的一位数字.在这样做的时候,我们可以预先检查向原整数附加另一位数字是否会导致溢出. 算法: 反转整数的方法可以与反转字符串进行类比. 我们想重复"弹出" x 的最后一位数字,并将它"推入"到 rev 的后面.最后,rev 将与 x 相反. 要在没有辅助堆栈 / 数组的帮助下…