一、题目描述 Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e.,0 1 2 4 5 6 7might become4 5 6 7 0 1 2). Find the minimum element. You may assume no duplicate exists in the array. 二、分析 这题难度有,因为他默认的是数组中所有的元素是不同的。只...
我们须要分情况处理。当然也能够无论这些直接遍历,即从头到尾扫描,找出最小的数;假设依据题意,我们须要考虑 1、 当严格单调时,因为是排好序的所以我们直接return 第一个元素就可以;或者直接利用二分法查找也能够。 2、 当有拐点时,我们就能够二分查找高速找到最小值。 综上所诉。考虑两种情况能够提高效率,可是直...
LeetCode 153. Find Minimum in Rotated Sorted Array 程序员木子 香港浸会大学 数据分析与人工智能硕士在读 来自专栏 · LeetCode Description 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...
Find the minimum element. The array may contain duplicates. 【分析】 这一道题是上一道题的拓展延伸:[LeetCode]153.Find Minimum in Rotated LeetCode33.搜索旋转排序数组 题目来源: https://leetcode-cn.com/problems/search-in-rotated-sorted-array/ 题目描述: 代码如下:...
Can you solve this real interview question? Find Minimum in Rotated Sorted Array - Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become: * [4,5,6,7,0,1,2] if
LeetCode154:Find Minimum in Rotated Sorted Array II 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]). Find the minimum element. The arra......
Leetcode - Find Minimum in Rotated Sorted Array II [分析] Find Minimum in Rotated Sorted Array是基础版,可用于打开基础思路,升级版是要考虑数组可能有重复元素。将题目中的case旋转多次找出思路,直接的思路是顺序遍历找到第一个非升序的元素,这种做法复杂度是O(N),优化的做法是每次通过比较中间元素和左边界...
Leetcode Find Minimum in Rotated Sorted Array 题目大意: 对一个有序数组翻转, 就是随机取前K个数,移动到数组的后面,然后让你找出最小的那个数,注意,K有可能是0,也就是没有翻转。 毫无疑问,遍历一次肯定可以找到,但这样时间复杂度是O(n),如果你在面试的时候遇到这样的问题,你这样回答面试官肯定不会满意的...
Suppose a sorted array is rotated at some pivot unknown to you beforehand. 0 1 2 4 5 6 7might become4 5 6 7 0 1 2). Find the minimum element. You may assume no duplicate exists in the array. 思路: 有序数组旋转后,如果mid元素比low大,则左边有序,右边乱序,考虑最左元素是否最小元素...
Leetcode Find Minimum in Rotated Sorted Array 题目大意: 对一个有序数组翻转, 就是随机取前K个数,移动到数组的后面,然后让你找出最小的那个数,注意,K有可能是0,也就是没有翻转。 毫无疑问,遍历一次肯定可以找到,但这样时间复杂度是O(n),如果你在面试的时候遇到这样的问题,你这样回答面试官肯定不会满意的...