Search in rotated sorted array of leetcode, which refers to the minimum number of the rotation array of Offer

Search in rotated sorted Array

Suppose a sorted array is rotated at some unknown to you beforehand.

(I. e .,0 1 2 4 5 6 7Might become4 5 6 7 0 1 2).

You are given a target value to search. If found in the array return its index, otherwise return-1.

You may assume no duplicate exists in the array.

Class solution {public: int search (int A [], int N, int target) {int left = 0, Right = n-1; while (left <= right) {int mid = left + (right-left)> 1); if (a [Mid] = target) return mid; if (A [Mid]> A [right]) // The first increment interval {if (a [left] <= target & target <A [Mid]) right = mid-1; else left = Mid + 1;} else // The second increment interval {if (a [Mid] <target & target <= A [right]) left = Mid + 1; else right = mid-1 ;}} return-1 ;}};

Search in rotated sorted array II

Follow up for "search in rotated sorted array ":
What if duplicates are allowed?

Wocould this affect the run-time complexity? How and why?

Write a function to determine if a given target is in the array.

Class solution {public: bool search (int A [], int N, int target) {int left = 0, Right = n-1; while (left <= right) {int mid = left + (right-left)> 1); if (a [Mid] = target) return true; if (A [Mid] = A [right]) // traverse the last section {for (; Right> = mid; -- right) {if (a [right] = target) return true;} else if (a [Mid]> A [right]) // The first increment interval {if (a [left] <= target & target <A [Mid]) Right = mid-1; else left = Mid + 1 ;} else // The second increment interval {if (a [Mid] <target & target <= A [right]) Left = Mid + 1; else right = mid-1 ;}} return false ;}};

Sword refers to offer: Moving the first several elements of an array to the end of an array, which is called the rotation of an array. Input a rotation of an incremental sorting array and output the smallest element of the rotating array. For example, if an array {3, 4, 5, 1, 2} is a rotation of {1, 2, 3, 4, 5}, the minimum value of this array is 1.

Train of Thought: here we provide two similar ideas, which are not the same as those on the offoffoffer. I think the offoffoffis hard to understand, but the efficiency of this algorithm is not as high as that in the book, the nine-degree space has timed out.

Int rotatearray (int * data, int N) {int left = 0, Right = n-1, minvalue = int_max; while (left <= right) {int mid = left + (right-left)> 1); If (data [Mid] <minvalue) minvalue = data [Mid]; if (data [Mid] = data [right]) // linear traversal {for (; Right> = mid; right --) {If (minvalue> data [right]) minvalue = data [right] ;}} else if (data [Mid]> data [right]) Left = Mid + 1; // The first increment interval else right = mid-1; // The second increment interval} return minvalue ;}

Search in rotated sorted array of leetcode, which refers to the minimum number of the rotation array of Offer