rotated专题

LeetCode - 33. Search in Rotated Sorted Array

33. Search in Rotated Sorted Array Problem's Link ---------------------------------------------------------------------------- Mean: 给你一个数组，这个数组是由两个有序的数组拼接成的（无重复元素），在这个数组中查找元素k的下标. anal

LeetCoder 33. Search in Rotated Sorted Array(二分)

是把一个升序的数组，前面的连续的一部分（可以是空）放在数组的后面，然后再这个数组这中找一个给出的值，数组中不存在重复的元素。这个题目也是《剑指offer》二分查找的一道例题 class Solution {public:int search(vector<int>& nums, int target) {int l = 0, r = nums.size()-1;while(l <= r)

leetcode 刷题之路 35 Search in Rotated Sorted Array II （旋转排序数组查找终极版）

Follow up for "Search in Rotated Sorted Array": What if duplicates are allowed? Would this affect the run-time complexity? How and why? Write a function to determine if a given target is in the a

Leetcode103: Search in Rotated Sorted Array II

Follow up for "Search in Rotated Sorted Array": What if duplicates are allowed? Would this affect the run-time complexity? How and why? Write a function to determine if a given target is in the a

LeetCode | Find Minimum in Rotated Sorted Array

Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). Find the minimum element. You may assume no duplicate exists in the arra

旋转数组的最小数字 154. Find Minimum in Rotated Sorted Array II

把一个数组最开始的若干个元素搬到数组的末尾，我们称之为数组的旋转。输入一个非递减排序的数组的一个旋转，输出旋转数组的最小元素。例如数组{3,4,5,1,2}为{1,2,3,4,5}的一个旋转，该数组的最小值为1。 NOTE：给出的所有元素都大于0，若数组大小为0，请返回0。几个Corner case： 1.如果数组实际没有反转是完全有序的 2.如果找到的mid 和star

LeetCode第33题之 Search in Rotated Sorted Array

C++代码： #include <iostream>#include <vector>using namespace std;class Solution {public:int search(vector<int>& nums, int target) {int sz = nums.size();int l = 0;int r = sz-1;int mid;while (l<=r){mi

leetcode-33. Search in Rotated Sorted Array

leetcode-33. Search in Rotated Sorted Array 题目： Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). You are given a target v

leetcode No81. Search in Rotated Sorted Array II

Question： Follow up for "Search in Rotated Sorted Array": What if duplicates are allowed? Would this affect the run-time complexity? How and why? Write a function to determine if a given target

*Leetcode 81. Search in Rotated Sorted Array II

常规容易想到的解法：基本就是两点：1、判断在两个上升的段的哪个段 2、mid跟r相等的时候r--可以解决问题的 class Solution {public:bool search(vector<int>& nums, int target) {if(nums.size() == 0) return false;int l = 0, r = nums.size()-1;//int

leetcode 33. Search in Rotated Sorted Array(二分查找）

只要找到枢轴，在左边或者右边查找target，因为是有序的，所以使用二分查找，这里实现了递归和非递归两种二分查找： /*int binarySearch(int *nums,int left,int right,int target) //递归的{int mid;if(left>=right){if(nums[left]==target)return left;else if(n

《leetCode》：Search in Rotated Sorted Array II

题目 Follow up for "Search in Rotated Sorted Array":What if duplicates are allowed?Would this affect the run-time complexity? How and why?Write a function to determine if a given target is in the arra

《leetCode》：Search in Rotated Sorted Array

题目 Suppose a sorted array is rotated at some pivot unknown to you beforehand.(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).You are given a target value to search. If found in the array return its

[总结] orb(Oriented Fast and rotated Brief) 1.详解

[总结] orb(Oriented FAST and Rotated BRIEF) 2.代码

参考： ORB 特征提取算法（实践篇一） - Alex777 - 博客园参考：【OpenCV学习笔记】之图像金字塔（Image Pyramid）_zhu_hongji的博客-CSDN博客_图像金字塔 1. 引言 1.1 图像金字塔（Image Pyramid）尺度调整，可以用 resize

[LeetCode] 33. Search in Rotated Sorted Array @ python

一．题目：有一个不含重复数字的升序数组，可能做了一些旋转操作，给定一个数查询是否存在这个数，若存在返回下标否则返回-1.时间复杂度为O(log n). Example 1: Input: nums = [4,5,6,7,0,1,2], target = 0Output: 4 Example 2: Input: nums = [4,5,6,7,0,1,2], target = 3Out

LeetCode 154 Find Minimum in Rotated Sorted Array II (二分或分治)

Follow up for "Find Minimum in Rotated Sorted Array": What if duplicates are allowed? Would this affect the run-time complexity? How and why? Suppose a sorted array is rotated at some pivot unknow

leetcode 84 Search in Rotated Sorted Array II 二分循环变种

题目 Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]). You are given a target value to search. If fo

LeetCode Search in Rotated Sorted Array

题目： Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). You are given a target value to search. If found in the array r

[LeetCode]81.Search in Rotated Sorted Array II

【题目】 Search in Rotated Sorted Array II Total Accepted: 3749 Total Submissions: 12937 My Submissions Follow up for "Search in Rotated Sorted Array": What if duplicates are allowed? Would

[LeetCode]33.Search in Rotated Sorted Array

【题目】 Search in Rotated Sorted Array Total Accepted: 5827 Total Submissions: 20925 My Submissions Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2

leetcode 081 Search in Rotated Sorted Array II

Follow up for "Search in Rotated Sorted Array": What if duplicates are allowed? Would this affect the run-time complexity? How and why? Write a function to determine if a given target is in

LeetCode OJ:Search in Rotated Sorted Array II

Search in Rotated Sorted Array II Follow up for "Search in Rotated Sorted Array": What if duplicates are allowed? Would this affect the run-time complexity? How and why? Write a function t

leetcode-153 Find Minimum in Rotated Sorted Array

剑指offer上的题目，旋转数组的最小值利用二分查找的思想，但是和二分查找又有不同 <span style="font-family:Microsoft YaHei;font-size:14px;">int findMin(int num[], int n) {int low = 0,high = n-1;while(low < high){if(num[low] < num[high]

153. Find Minimum in Rotated Sorted Array 旋转数组的最小数字

题目描述把一个数组最开始的若干个元素搬到数组的末尾，我们称之为数组的旋转。输入一个非递减序列的一个旋转，输出旋转数组的最小元素。例如数组{3,4,5,1,2}为{1,2,3,4,5}的一个旋转，该数组的最小值为1。解答　　最直观的解法从头到尾遍历整个数组，找到最小元素。这个解法的时间复杂度是O(n)。　　注意到，旋转之后的数组实际上可以划分为两个排序的子数组，且前面的子数组的元素都

Leetcode C++ 33. Search in Rotated Sorted Array

Leetcode 33. Search in Rotated Sorted Array 原文： Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).