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.…
题目链接 :https://leetcode.com/problems/search-in-rotated-sorted-array/?tab=Description   Problem :当前的数组是一个经过排序之后的循环有序数组,但是该数组的主元选择并不一定是下标为i=0的第一个元素. 例如有序数组为:{1,2,3,4,5,6}  其循环有序数组可能为: {3,4,5,6,1,2}. {6,1,2,3,4,5}.{4,5,6,1,2,3}等   进行查询操作,使用折半查找. 需要不断判断nu…
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…
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…
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.…
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. You…
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 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 retu…
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 index, otherwise return -1. You may assume no duplic…
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[…