一句话思路:反正只是寻找一个最小区间,断开也能二分.根据m第一次的落点,来分情况讨论. 一刷报错: 结构上有根本性错误:应该是while里面包括if,不然会把代码重复写两遍,不好. //situation1 if (nums[mid] > nums[start]) { while (start + 1 < mid) { mid = start + (end - start) / 2; if (nums[mid] == target) { return mid; } else if (nums[…

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]). You are given a target value to search. If found in the array return its index, otherwise return -1.…

33. Search in Rotated Sorted Array 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). You are given a target value to search. If found in the array return i…

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…

leetcode - 33. Search in Rotated Sorted Array - Medium descrition 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). You are given a target value to search.…

LeetCode 33 Search in Rotated Sorted Array [binary search] <c++> 给出排序好的一维无重复元素的数组,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的下标,如果没有,返回-1.时间复杂度限制\(O(log_2n)\) C++ 我的想法是先找到数组中最大值的位置.然后以此位置将数组一分为二,然后在左右两部分分别寻找target. 二分寻找最大值的时候,因为左半部分的数一定大于nums[l],所以n…

# -*- coding: utf8 -*-'''__author__ = 'dabay.wang@gmail.com' 33: Search in Rotated Sorted Arrayhttps://oj.leetcode.com/problems/search-in-rotated-sorted-array/ Suppose a sorted array is rotated at some pivot unknown to you beforehand.(i.e., 0 1 2 4 5…

33. Search in Rotated Sorted Array Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e., [,,,,,,] might become [,,,,,,]). You are given a target value to search. If found . You may assume no duplicate e…

一.题目说明 这个题目是33. Search in Rotated Sorted Array,说的是在一个"扭转"的有序列表中,查找一个元素,时间复杂度O(logn). 二.我的解答 这是一个查找,根据复杂度,我们知道只能用二分查找.但由于这个不是一个完全的有序列表,故需要改造.先写出二分查找(来源于数据结构): 二分查找: int search(vector<int>& nums, int target){ if(nums.size()<0) return…

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]). You are given a target value to search. If found in the array return its index, otherwise return -1.…