标签:Binary 153 nums rotated Rotated mid times array was
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 it was rotated4
times.[0,1,2,4,5,6,7]
if it was rotated7
times.
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.
You must write an algorithm that runs in O(log n) time.
Example 1:
Input: nums = [3,4,5,1,2] Output: 1 Explanation: The original array was [1,2,3,4,5] rotated 3 times.
Example 2:
Input: nums = [4,5,6,7,0,1,2] Output: 0 Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.
Example 3:
Input: nums = [11,13,15,17] Output: 11 Explanation: The original array was [11,13,15,17] and it was rotated 4 times.
Constraints:
n == nums.length
1 <= n <= 5000
-5000 <= nums[i] <= 5000
- All the integers of
nums
are unique. nums
is sorted and rotated between1
andn
times.
这题跟LeetCode 33. Search in Rotated Sorted Array类似,解题思路也基本一样用二分查找法,把数组分成左右两个区间后判断最小值是落在左半区间还是右半区间,关键点是判断的条件该如何确定呢?
如果把中点值与最左边值比较即nums[mid] > nums[left],左半区间是递增的,但是最小值可能在左半区间也可能在右半区间。如果把中点值与最右边值比较即nums[mid] < nums[right],右半区间是递增的,最小值可以确定一定在左半区间(包含nums[mid])。因此本题以nums[mid]与nums[right]比较为判断条件。
如果nums[mid]小于nums[r],最小值落在左半区间(包含nums[mid]),所以右边界更新为r=mid。那如果nums[mid]不小于nums[r],最小值就落在了右半区间(不包含nums[mid]),所以左边界更新为l=mid + 1。按此条件一直循环折半查找直到区间只剩一个数,这个数就是最小值。
class Solution:
def findMin(self, nums: List[int]) -> int:
l, r = 0, len(nums) - 1
while l < r:
mid = l + (r - l) // 2
if nums[mid] < nums[r]:
r = mid
else:
l = mid + 1
return nums[l]
标签:Binary,153,nums,rotated,Rotated,mid,times,array,was 来源: https://blog.csdn.net/hgq522/article/details/122020280
本站声明: 1. iCode9 技术分享网(下文简称本站)提供的所有内容,仅供技术学习、探讨和分享; 2. 关于本站的所有留言、评论、转载及引用,纯属内容发起人的个人观点,与本站观点和立场无关; 3. 关于本站的所有言论和文字,纯属内容发起人的个人观点,与本站观点和立场无关; 4. 本站文章均是网友提供,不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属;如您发现该文章侵犯了您的权益,可联系我们第一时间进行删除; 5. 本站为非盈利性的个人网站,所有内容不会用来进行牟利,也不会利用任何形式的广告来间接获益,纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。