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,遍历一次肯定能够找到,但这样时间复杂度是O(n)。假设你在面试的时候遇到这种问题,你这样回答面试官肯定不会惬意的。
mid 和start 比较 mid > start : 最小值在左半部分。 mid < start: 最小值在左半部分。 无论大于还是小于,最小值都在左半部分,所以 mid 和start 比较是不可取的。 mid 和end 比较 mid < end:最小值在左半部分。 mid > end:最小值在右半部分。 所以我们只需要把 mid 和end 比较,mid < end 丢...
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
153. Find Minimum 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]). Find the min element. You may assume no duplicate exists in the array. Example ...
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. You may assume no duplicate exists in the array. ...
主要思路还是跟Search in Rotated Sorted Array差点儿相同。还是通过左边界和中间的大小关系来得到左边或者右边有序的信息。假设左半边有序。那么左半边最小就是左边第一个元素,能够和当前最小相比取小的,然后走向右半边。否则,那么就是右半半边第一个元素,然后走向左半边。这样子每次能够截掉一半元素,所以最后复杂...
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 [4,5,6,7,0,1,2]). Find the minimum element. The array m…
Search in rotated sorted array - Leetcode 33 - Python 呼吸的chou 0 0 Code-It-Yourself! Tetris - Programming from Scratch (Quick and Simple C++) 呼吸的chou 4 0 Decode Ways - Dynamic Programming - Leetcode 91 - Python 呼吸的chou 0 0 ...
Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]]. Given the sorted rotated array nums of unique elements, return the minimum element of this array. ...
153. Find Minimum in Rotated Sorted Array 没有重复的情况。当然要用二分查找,我永远也忘不了头条第一挂,竟然是这么简单的题没整明白。 classSolution(object):deffindMin(self,nums):""" :type nums: List[int] :rtype: int """iflen(nums)==1:returnnums[0]l,r=0,len...