目录 unordered系列关联式容器底层结构模拟实现 1. unordered系列关联式容器 在c++98中,STL提供了底层为红黑树结构的一系列关联式容器,在查询时效率可达到 l o g 2 N log_2N log2N,即最差情况下需要比较红黑树的高度次,当树中的结点非常多时,查询效率也不理想。最好的查询是,进行很少的比较次数就能将元素找到,因此在c++11中,stl又提供了4个un
Problem: 2748. 美丽下标对的数目 👨🏫 参考题解 🍻 暴力法 class Solution {public int countBeautifulPairs(int[] nums) {int res = 0; int n = nums.length;for(int i = 0; i < n; i++){while(nums[i] >= 10){nums[i]
原题: /*** Created by gouthamvidyapradhan on 10/03/2017.* Given two strings s and t, write a function to determine if t is an anagram of s.* <p>* For example,* s = "anagram", t = "nagaram", return true
原题: /*** Created by gouthamvidyapradhan on 09/03/2017.* Given an array of integers, return indices of the two numbers such that they add up to a specific target.* <p>* You may assume that each input
原题: /*** Created by gouthamvidyapradhan on 25/03/2017.* Given a string, sort it in decreasing order based on the frequency of characters.* <p>* Example 1:* <p>* Input:* "tree"* <p>* Output:* "eert"*
原题: /*** Created by gouthamvidyapradhan on 18/10/2017.* Given an array nums and a target value k, find the maximum length of a subarray that sums to k. If there isn't* one, return 0 instead.Note:The
原题: /*** Created by gouthamvidyapradhan on 28/03/2017.* Given an array of integers and an integer k, you need to find the number of unique k-diff pairs in the array. Here a k-diff pair is defined as
原题: /*** Created by gouthamvidyapradhan on 10/03/2017.* Given an array of strings, group anagrams together.* <p>* For example, given: ["eat", "tea", "tan", "ate", "nat", "bat"],* Return:* <p>* [* ["a
InnoDB存储引擎监测到同样的二级索引不断被使用,那么它会根据这个二级索引,在内存上根据二级索引树(B+树)上的二级索引值,在内存上构建一个哈希索引,来加速搜索。 查看是否开启自适应哈希索引 show variables like 'innodb_adaptive_hash_index'; 查看自适应哈希索引分区 show variables like 'innodb_adaptive