Search in Rotated Sorted Array II

search.java

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

(i.e., 0 1 2 4 5 6 7 might become 4 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.

Algorithm:
1. Based on the algorithm from Search in Rotated Sorted Array I, by adding
if A[mid] == A[right], and then decrease right--;
else if they are different, then follow the origional algorithm to move 
the pointer.
*/
public class Solution {
    public boolean search(int[] A, int target) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int left = 0;
        int right = A.length - 1;
        while(right >= left) {
            int mid = (right + left) / 2;
            if(A[mid] > A[right]) {
                if(A[left] <= target && target < A[mid]) {
                    right = mid - 1;
                } else if(A[mid] == target) {
                    return true;
                } else {
                    left = mid + 1;
                }
            } else if(A[mid] < A[right]) {
                if(A[mid] < target && target <= A[right]) {
                    left = mid + 1;
                } else if(target == A[mid]) {
                    return true;
                } else {
                    right = mid - 1;
                }
            } else if(A[mid] == target) {
                return true;
            } else {
                right--;
            }
        }
        return false;
    }
}