LeetCode OJ - Search in Rotated Sorted Array II

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.

分析：二分法的基礎之上去重

class Solution {
public:
    bool search(int A[], int n, int target) {
        if(n == 0) return -1;
        
        int left = 0, right = n - 1;
        while(left <= right) {
            while(left < right && A[left] == A[left + 1]) left++;
            while(left < right && A[right] == A[right - 1]) right--;
            
            int mid = (left + right) >> 1;
            if(target == A[mid]) {
                return true;
            } else if(A[left] > A[mid]) {
            //情況一: 有序區在[mid, rihgt]
                if(A[mid] < target && target <= A[right]) {
                    left = mid + 1;
                } else {
                    right = mid - 1;
                }
            } else {
            //情況二: 有序區在[left, mid]
                if(A[left] <= target && target < A[mid]) {
                    right = mid - 1;
                } else {
                    left = mid + 1;
                }
            }
        }
        
        return false;        
    }
};