Quick Sort: Time complexity: best case O(n*lgn), worst case O(n^2) Space complexity: Best case O(lgn) -> call stack height Worse case O(n^2) -> call stack height Merge Sort Time complexity: always O(n*lgn) because we always divide the array in halve
时间&空间复杂度 时间复杂度: 通俗来说就是随着数据量的增加,程序运行的时间花费量是怎么变化的,时间复杂度常用大o表示.举个例子,猜数字,猜10个,100个.1000个,猜数的数据量是在增加的,但是实际运行程序花费的时间是怎么变化的呢,是线性的?常数的?还是指数的,它反映的是一个趋势.简而言之,时间复杂度反映的是程序执行时间随输入规模增长的量级.从某种程度上将可以展示算法的scale ability. how to analyze? e.g1:如下,当我们不断增长i的数值,循环的次数也跟着变化
Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array. Example: For num = 5 you should return [0,1,1,2,1,2]. Follow up: It is very
Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array. For example, Given nums = [0, 1, 3] return 2. Note: Your algorithm should run in linear runtime complexity. Could you implement it u
How to solve this HARD issue 1. Problem: A non-empty zero-indexed array A consisting of N integers is given. A peak is an array element which is larger than its neighbours. More precisely, it is an index P such that 0 < P < N − 1 and A[P − 1] < A
这几天把之前的设计模式回顾了一遍,整理了一点以前的项目.同学说,打算刷leetcode题目,也勾起了我的兴趣,索性也刷一些题目,再提高一些内功.刚开始进去,leetcode随机分配的题目,直接也就做了,在后来,按顺序做,现在做到了第五题.大概做了如下题目: 1. Two Sum Given an array of integers, find two numbers such that they add up to a specific target number. The function t
463. Island Perimeterhttps://leetcode.com/problems/island-perimeter/就是逐一遍历所有的cell,用分离的cell总的的边数减去重叠的边的数目即可.在查找重叠的边的数目的时候有一点小技巧,就是沿着其中两个方向就好,这种题目都有类似的规律,就是可以沿着上三角或者下三角形的方向来做.一刷一次ac,但是还没开始注意codestyle的问题,需要再刷一遍. class Solution { public: int islandPerime
Given an array of n integers where n > 1, nums, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i]. Solve it without division and in O(n). For example, given [1,2,3,4], return [24,12,8,6]. 题目
题目: Given a string array words, find the maximum value of length(word[i]) * length(word[j]) where the two words do not share common letters. You may assume that each word will contain only lower case letters. If no such two words exist, return 0. Exa
题目: Given a string that contains only digits 0-9 and a target value, return all possibilities to add binary operators (not unary) +, -, or * between the digits so they evaluate to the target value. Examples: "123", 6 -> ["1+2+3", &q
题目: Design an algorithm to encode a list of strings to a string. The encoded string is then sent over the network and is decoded back to the original list of strings. Machine 1 (sender) has the function: string encode(vector<string> strs) { // ... y
