LeetCode 81 Search in Rotated Sorted Array II [binary search] <c++> 给出排序好的一维有重复元素的数组,随机取一个位置断开,把前半部分接到后半部分后面,得到一个新数组,在新数组中查找给定数是否存在,时间复杂度限制\(O(log_2n)\) C++ 因为有重复元素存在,nums[l] <= nums[mid]不能说明[l,mid]区间内一定是单调的,比如数组[1,2,3,1,1,1,1],但是严格小于和严格大于的情况还是可以…

This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates. 思路 该题是[leetcode]33. Search in Rotated Sorted Array旋转过有序数组里找目标值 的followup 唯一区别是加了line24-26的else语句来skip duplicates 代码 class Solution { public boolean sear…

Follow up for "Search in Rotated Sorted Array":What if duplicates are allowed? Would this affect the run-time complexity? How and why? Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4…

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 found in the array return true, otherwise return false. Ex…

Follow up for "Search in Rotated Sorted Array":What if duplicates are allowed? Would this affect the run-time complexity? How and why? Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4…

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 array. 思路:此题在解的时候,才发现Search in Rotated Sorted Array…

题目: 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). Write a function to determine if a given target is in the array. The array may contain duplicates. 思路…

题目链接:https://leetcode.com/problems/search-in-rotated-sorted-array-ii/#/description   姊妹篇:http://www.cnblogs.com/zpfbuaa/p/6531773.html   给出一个循环有序数组,与其姊妹篇不同的是其中允许元素出现重复多次.给定一个数字target,查找该数字是否存在于该数组中.   姊妹篇中的解决方法是使用折半查找,通过判断nums[left]和nums[mid]以及nums[r…

原题地址 如果不存在重复元素,仅通过判断数组的首尾元素即可判断数组是否连续,但是有重复元素的话就不行了,最坏情况下所有元素都一样,此时只能通过线性扫描确定是否连续. 设对于规模为n的问题的工作量为T(n),则有T(n) = T(n/2) + O(n),根据主定理,可以求得T(n) = O(n).和之前的O(logn)相比还是退化了不少. 代码: bool monop(int A[], int l, int r) { while (l < r && A[l] <= A[r]) l…

这4个题都是针对旋转的排序数组.其中153.154是在旋转的排序数组中找最小值,33.81是在旋转的排序数组中找一个固定的值.且153和33都是没有重复数值的数组,154.81都是针对各自问题的版本1增加了有重复数值的限制条件. 153.154都需要考虑是否旋转成和原数组一样的情况,特别的,154题还需要考虑10111和11101这种特殊情况无法使用二分查找. 33.81则不用考虑这些情况. 找最小值的题主要是利用最小值小于他前一个位置的数,同时也小于后一个位置的数 153. Find Mini…