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package Algorithms.array;

/*

* https://oj.leetcode.com/problems/search-in-rotated-sorted-array-ii/

*

* Search in Rotated Sorted Array II Total Accepted: 18427 Total Submissions: 59728 My Submissions

Follow up for "Search in Rotated Sorted Array":

What if duplicates are allowed?

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

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

* */

public class SearchInRotatedSortedArray2 {

/*

In this function, the complex may go to O(n). It depents on how many duplicates exit in the array.

if there are M duplicates, the complexity may be O(logN + M).

*/

public boolean search(int[] A, int target) {

if (A == null || A.length == 0) {

return false;

}

// setup two point to the left and right of the array.

int l = 0, r = A.length - 1;

while (l <= r) {

// get the mid point.

int mid = l + (r - l)/2;

if (target == A[mid]) {

return true;

}

if (A[mid] > A[l]) {

// the left side is sorted.

if (target <= A[mid] && target >= A[l]) {

// target is in the left side.

r = mid - 1;

} else {

l = mid + 1;

}

} else if (A[mid] < A[l]) {

// the right side is sorted.

if (target <= A[r] && target >= A[mid]) {

// target is in the right side.

l = mid + 1;

} else {

r = mid - 1;

}

} else {

// when A[mid] == A[l], we can't determin, just move the

// left point one

l++;

}

}

return false;

}

}