Search in Rotated Sorted Array II

O(N)

if element can repeat, the worst case, you cannot throw away any section. eg. [1, 1, 1, 1, 0, 1, 1] target = 0, you cannot throw any section. We have to loop through to check. 

public class Solution {
    /** 
     * param A : an integer ratated sorted array and duplicates are allowed
     * param target :  an integer to be search
     * return : a boolean 
     */
    public boolean search(int[] A, int target) {
        // write your code here
        if (A == null || A.length == 0) {
            return false;
        }
        
        for (int i = 0; i < A.length; i++) {
            if (A[i] == target) {
                return true;
            } 
        }
        return false;
    }
}