#include <string> #include <iostream> //查询性能最高 //允许重复的,hash_map #include <unordered_map> #include <algorithm> using namespace std; void main() { //允许重复的映射 unordered_multimap<},{ },{ } }; mymap.insert(pair<)); mymap.insert(pai

Design a data structure that supports all following operations in average O(1) time. Note: Duplicate elements are allowed. insert(val): Inserts an item val to the collection. remove(val): Removes an item val from the collection if present. getRandom:

Given an array of integers and an integer k, return true if and only if there are two distinct indices i and j in the array such that nums[i] = nums[j] and the difference between i and j is at most k. (Old Version) Given an array of integers and an i

Given an array of integers, find if the array contains any duplicates. Your function should return true if any value appears at least twice in the array, and it should return false if every element is distinct. 这道题不算难题,就是使用一个哈希表,遍历整个数组,如果哈希表里存在,返回fal

All DNA is composed of a series of nucleotides abbreviated as A, C, G, and T, for example: "ACGAATTCCG". When studying DNA, it is sometimes useful to identify repeated sequences within the DNA. Write a function to find all the 10-letter-long seq

Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. For "bbbbb" the longest subst

转自:http://blog.csdn.net/truexf/article/details/17303263 一.vector vector采用一段连续的内存来存储其元素,向vector添加元素的时候,如果容量不足,vector便会重新malloc一段更大的内存,然后把原内存中的数据memcpy到新的内存中,并free原内存块,然后将新元素加入.vector的元素插入性能跟以下几个要素关系重大: 1. 插入的位置 头部插入:将所有元素后移,然后将新元素插入 中间插入:将插入点后面的元素后移,然

题目意思:求字符串中,最长不重复连续子串 思路:使用hashmap,发现unordered_map会比map快,设置一个起始位置,计算长度时,去减起始位置的值 eg:a,b,c,d,e,c,b,a,e 0 1 2 3 4 4 0 6 5 3 4 4 3 8 再次出现c时,将start值置为map[c]+1=3,对于下一个b,因为map[b]<start,则直接map[b]=i class Solution { public: int lengthOfLongestSubstring(string

Given a binary tree, return all duplicate subtrees. For each kind of duplicate subtrees, you only need to return the root node of any oneof them. Two trees are duplicate if they have the same structure with same node values. Example 1: 1 / \ 2 3 / /

Given a list of directory info including directory path, and all the files with contents in this directory, you need to find out all the groups of duplicate files in the file system in terms of their paths. A group of duplicate files consists of at l

Given a string, find the length of the longest substring without repeating characters. Example 1: Input: "abcabcbb" Output: 3 Explanation: The answer is "abc", with the length of 3. Example 2: Input: "bbbbb" Output: 1 Explana

Medium! 题目描述: 给定一个字符串,找出不含有重复字符的 最长子串 的长度. 示例: 给定 "abcabcbb" ,没有重复字符的最长子串是 "abc" ,那么长度就是3. 给定 "bbbbb" ,最长的子串就是 "b" ,长度是1. 给定 "pwwkew" ,最长子串是 "wke" ,长度是3.请注意答案必须是一个子串,"pwke" 是 子序列 而不是子串

map /multimap map是STL里重要容器之一. 它的特性总结来讲就是:所有元素都会根据元素的键值key自动排序(也可根据自定义的仿函数进行自定义排序),其中的每个元素都是<key, value>的键值对,map中不允许有键值相同的元素, 因此map中元素的键值key不能修改,但是可以通过key修改与其对应的value.如果一定要修改与value对应的键值key,可将已存在的key删除掉,然后重新插入. 定义原型: 它作用应用场景可用作 ①字典    ②统计次数 相关操作 (1)插入

今天是leetcode第一天,但是不太顺利.做这些,想不到 原题目: 我给的答案: class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { vector<int> result; ; i <nums.size() ; i ++){ ; j <nums.size() ; j ++){ if( nums[i] + nums[j] == target){ re

unordered_map: 和 unorder_set 相似,该容器内部同样根据 hash value 把键值对存放到相应的 bucket(slot)中,根据单个 key 来访问 value 的速度很快. unordered_multimap: 操作和 unorder_map 相同,不同点是 key 可以重复.通过 it.first(it->first) 访问 key 或者 it.second(it->second) 访问 mapped value. Iterators(只有有序的容器才有

stl:即标准模板库,该库包含了诸多在计算机科学领域里所常用的基本数据结构和基本算法 六大组件: 容器.迭代器.算法.仿函数.空间配置器.迭代适配器 迭代器:迭代器(iterator)是一种抽象的设计理念,通过迭代器可以在不了解容器内部原理的情况下遍历容器.除此之外,STL中迭代器一个最重要的作用就是作为容器(vector,list等)与STL算法的粘结剂,只要容器提供迭代器的接口,同一套算法代码可以利用在完全不同的容器中,这是抽象思想的经典应用.迭代器是STL中行为类似指针的设计模式,它可以提

1.题目描述 Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1. Examples:  s = "leetcode" return 0.  s = "loveleetcode", return 2. Note: You may assume the string contain only

我现在在做一个叫<leetbook>的免费开源书项目,力求提供最易懂的中文思路,目前把解题思路都同步更新到gitbook上了,需要的同学可以去看看 书的地址:https://hk029.gitbooks.io/leetbook/ 003. Longest Substring Without Repeating Characters[M] Longest Substring Without Repeating CharactersM 题目 思路 代码 题目 Given a string, fin

通过把未访问的结点放到unordered_map中来判断是否重复,代码如下: class Solution { public: int lengthOfLongestSubstring(string s) { if(s == "") ; unordered_map<char,int> hash; ; ; ;i < s.length();i ++) { if(hash.find(s[i]) != hash.end()) { i = hash[s[i]]; hash.cl

map的特性是,所有元素都会根据元素的键值自动被排序.map的所有元素都是pair,同时拥有键值(key)和实值(value).pair的第一元素被视为键值,第二元素被视为实值.map不允许两个元素拥有相同的键值multimap的特性以及用法与map完全相同,唯一的差别在于它允许键值重复.unordered_map与map的区别就在于不会根据key的大小进行排序. 1.插入数据的方法 #include <map> #include <string> #include <ios