{ "data": { "question": { "questionId": "153", "questionFrontendId": "153", "categoryTitle": "Algorithms", "boundTopicId": 1403, "title": "Find Minimum in Rotated Sorted Array", "titleSlug": "find-minimum-in-rotated-sorted-array", "content": "

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:

\n\n\n\n

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]].

\n\n

Given the sorted rotated array nums of unique elements, return the minimum element of this array.

\n\n

You must write an algorithm that runs in O(log n) time.

\n\n

 

\n

Example 1:

\n\n
\nInput: nums = [3,4,5,1,2]\nOutput: 1\nExplanation: The original array was [1,2,3,4,5] rotated 3 times.\n
\n\n

Example 2:

\n\n
\nInput: nums = [4,5,6,7,0,1,2]\nOutput: 0\nExplanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.\n
\n\n

Example 3:

\n\n
\nInput: nums = [11,13,15,17]\nOutput: 11\nExplanation: The original array was [11,13,15,17] and it was rotated 4 times. \n
\n\n

 

\n

Constraints:

\n\n\n", "translatedTitle": "寻找旋转排序数组中的最小值", "translatedContent": "已知一个长度为 n 的数组,预先按照升序排列,经由 1n旋转 后,得到输入数组。例如,原数组 nums = [0,1,2,4,5,6,7] 在变化后可能得到:\n\n\n

注意,数组 [a[0], a[1], a[2], ..., a[n-1]] 旋转一次 的结果为数组 [a[n-1], a[0], a[1], a[2], ..., a[n-2]]

\n\n

给你一个元素值 互不相同 的数组 nums ,它原来是一个升序排列的数组,并按上述情形进行了多次旋转。请你找出并返回数组中的 最小元素

\n\n

你必须设计一个时间复杂度为 O(log n) 的算法解决此问题。

\n\n

 

\n\n

示例 1:

\n\n
\n输入:nums = [3,4,5,1,2]\n输出:1\n解释:原数组为 [1,2,3,4,5] ,旋转 3 次得到输入数组。\n
\n\n

示例 2:

\n\n
\n输入:nums = [4,5,6,7,0,1,2]\n输出:0\n解释:原数组为 [0,1,2,4,5,6,7] ,旋转 4 次得到输入数组。\n
\n\n

示例 3:

\n\n
\n输入:nums = [11,13,15,17]\n输出:11\n解释:原数组为 [11,13,15,17] ,旋转 4 次得到输入数组。\n
\n\n

 

\n\n

提示:

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Now that the array is rotated, there would be a point in the array where there is a small deflection from the increasing sequence. eg. The array would be something like [4, 5, 6, 7, 0, 1, 2].", "You can divide the search space into two and see which direction to go.\r\nCan you think of an algorithm which has O(logN) search complexity?", "
    \r\n
  1. All the elements to the left of inflection point > first element of the array.
  2. \r\n
  3. All the elements to the right of inflection point < first element of the array.
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