来源:力扣(LeetCode) 链接:81. Search in Rotated Sorted Array II 解答 解法1 // 时间复杂度O(n), 空间复杂度O(1) class Solution { public: bool search(vector<int>& nums, int target) { int lo = 0; int hi = nums.size(); while (lo != hi) { int mid = lo + (hi - lo) / 2...
// LeetCode, Search in Rotated Sorted Array II // 时间复杂度O(n),空间复杂度O(1) class Solution { public: bool search(const vector<int>& nums, int target) { int first = 0, last = nums.size(); while (first != last) { const int mid = first + (last - first) / 2; if (...
linear_search.rs sorting string ciphers.rs data_structures.rs dynamic_programming.rs general.rs graph.rs leetcode_solutions.rs lib.rs math.rs searching.rs sorting.rs string.rs tests .gitignore Cargo.toml README.mdBreadcrumbs rust_algorithms /src /searching / linear_search.rs L...
Leetcode 简单17 x的平方根 x的平方根: PHP 24ms:牛顿迭代法,公式为(n +x/n)/2 最优化理论与方法-牛顿迭代法 与x轴交点的横坐标,称x1为ζ的一次近似值。过点做曲线的切线,并求该切线与x轴交点的横坐标,称为ζ的二次近似值。重复以上过程,得ζ的近似值序列,其中称为ζ的次...的前面几项来寻找方程...
【leetcode】经典算法题-Counting Bits 题目描述: 给定一个数字n,统计0～n之间的数字二进制的1的个数,并用数组输出 例子: For num = 5 you should return [0,1,1,2,1,2]. 要求: 算法复杂复o(n) 空间复杂度o(n) 原文描述: Given a non negative integer number num. For every numbers i in ...
https://www.amazon.com/Introduction-Linear-Algebra-Gilbert-Strang/dp/0980232775/ Went through 85%-90% of this book on BART.. as my step 2 to learn Mac
// LeetCode, Search in Rotated Sorted Array // 时间复杂度O(log n),空间复杂度O(1) class Solution { public: int search(int A[], int n, int target) { int first = 0, last = n; while (first != last) { const int mid = first + (last - first) / 2; if (A[mid] == targe...