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This article mainly introduces "LeetCode problem solution how to achieve rotation array numbers". In daily operation, I believe that many people have doubts about how to achieve rotation array numbers in LeetCode problem solution. The editor consulted all kinds of data and sorted out simple and easy-to-use operation methods. I hope it will be helpful for you to answer the question of "how to achieve rotation array numbers in LeetCode problem solution". Next, please follow the editor to study!

Topic: minimum number of rotated array

Moving the first elements of an array to the end of the array is called the rotation of the array. Enter a rotation of an incrementally sorted array and output the smallest element of the rotation array. For example, the array [3pyrm 4rec 5je 1je 2] is a rotation of [1rect 2rec 3je 4je 5], with a minimum value of 1.

Example 1:

Input: [3pyrrine 4pyrrine 5pyrrine 1] output: 1exampl 2:

Input: [2, 2, 2, 2, 2, 0, 1] output: 0

Solution one

First of all, find the stem of the topic:

Incremental sort array (repeatable), rotation, minimum element

That is, an incremental array that moves a portion to the end of the array, such as:

[1] / / after rotation [3pyrmology, 4pyrrine, 5pjor2]

Find the smallest number.

So the first solution that we can easily think of is to traverse the array and find that when a number is smaller than the one before it, then this number is the smallest number we are looking for.

Because normally, the latter number is larger than the previous number, so there is less than the previous number, so it is the demarcation point of the rotation array, that is, the minimum number.

Class Solution {public int minArray (int [] numbers) {for (int iTuno [iTunes 1]) {return numbers [iTunes 1];}} return numbers [0];}}

Method consumption

I won't write this anymore. Because each test case is different, the result is too different, there is no reference.

Time complexity

Iterate through the array once, so the time complexity is O (n).

Space complexity

No other space is used, so the space complexity is O (1).

Solution two

Dichotomy.

Some people may wonder, isn't dichotomy used to find sequential arrays, and does it count after rotation?

In retrospect, the key point of dichotomy is:

If you take any key number, you can determine which range we want the value to be in (before and after the key number).

So can we do this in our rotation array?

For example, if we take the intermediate value an and the final value b, if an is greater than b, it means that the dividing line is between an and b, and the data before an is sorted correctly.

If an is less than b, the cut-off value is before a, and the data between an and b are sorted correctly.

For example, the median value of 5 is greater than the final value of 2, which means that the dividing line is between 5 and 2, that is, 1.

In the case of class Solution {public int minArray (int [] numbers) {int left=0; int right=numbers.length-1; / / dichotomy search condition while (leftnumbers [right]), mid cannot be minimum, so it is set to mid+1.

At this point, the study of "how to realize the number of rotating array in LeetCode" is over. I hope to be able to solve your doubts. The collocation of theory and practice can better help you learn, go and try it! If you want to continue to learn more related knowledge, please continue to follow the website, the editor will continue to work hard to bring you more practical articles!

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