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. The array may contain duplicates. 这道题目就是Find Minimum in Rotated Sorted Array的增强版,在数组中增加了重复的元素。 具体的解法...
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). Find the minimum element. The array may contain duplicates. 解法1:本题旋转数组最小值的扩展,这次数组中可以存在重复元素了。上题因数组中无重复元素,因此...
This is a follow up problem to Find Minimum in Rotated Sorted Array. Would allow duplicates affect the run-time complexity? How and why? 描述 假设按照升序排序的数组在预先未知的某个点上进行了旋转。 ( 例如,数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。 请找出其中最小的...
Follow up for “Find Minimum 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 5 6 7 might become 4 5 ...
This is a follow up problem toFind Minimum in Rotated Sorted Array. Would allow duplicates affect the run-time complexity? How and why? 题目大意 假设按照升序排序的数组在预先未知的某个点上进行了旋转。( 例如,数组 [0,1,2,4,5,6,7] 可能变为 [4,5,6,7,0,1,2] )。请找出其中最小的元...
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...
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 ...
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
Notice thatrotatingan 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 that may containduplicates, returnthe minimum element of this array. ...
接Find Minimum in Rotated Sorted Array, 假设有重复数字。 Solution 解法依然是二分搜索,不过多了一种情况要考虑。 classSolution(object):deffindMin(self,nums):""" :type nums: List[int] :rtype: int """L,R=0,len(nums)-1whileL<Randnums[L]>=nums[R]:M=(L+R)/2ifnums[M]>nums[R]:...