LeetCode刷题笔录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.

如果有可以重复元素的情况，那么有可能会不能判断哪个half是排好序的。比如数组13333， rotate成31333，A[l]和A[m]都是3。这时候只能让l往右移动直到A[l]!=A[m]为止。

最坏的情况，时间复杂度可能会变成O(n)。

public class Solution {
    public boolean search(int[] A, int target) {
        if(A == null || A.length == 0)
            return false;
        int l = 0;
        int r = A.length - 1;
        
        while(l <= r){
            int m = (l + r) / 2;
            if(A[m] == target)
                return true;
            //the left half is sorted
            if(A[l] < A[m]){
                if(A[l] <= target && A[m] > target)
                    r = m - 1;
                else
                    l = m + 1;
            }
            //the right half is sorted
            else if(A[l] > A[m]){
                if(A[m] < target && A[r] >= target)
                    l = m + 1;
                else
                    r = m - 1;
            }
            //cannot tell which half is sorted, must increment l until A[l] is not equal to A[m]
            else{
                l++;
            }
        }
        return false;
    }
}