Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]).

You are given a target value to search. If found in the array return true, otherwise return false.

Example 1:

Input: nums = [2,5,6,0,0,1,2]
 
, target = 0
Output: true

Example 2:

Input: nums = [2,5,6,0,0,1,2], target = 3
Output: false

Follow up:

• This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates.
• Would this affect the run-time complexity? How and why?

binary search

同33. Search in Rotated Sorted Array　　https://www.cnblogs.com/fatttcat/p/10063254.html

由於可以有重複元素，遇到 [1, 1, 3, 1] 情況無法判斷，需要增加一種情況：當nums[l] == nums[m]時，當前元素重複，l向右移動一位

時間：O(logN) , worst case O(N)，空間：O(1)

class Solution {
    public boolean search(int[] nums, int target) {
        if(nums == null
 
 || nums.length == 0) return false;
        int l = 0, r = nums.length - 1;
        
        while(l + 1 < r) {
            int m = l + (r - l) / 2;
            if(nums[m] == target) return true;
            if(nums[l] < nums[m]) {
                if(nums[l] <= target && target <= nums[m])
                    r = m;
                else
                    l = m;
            }
            else if(nums[l] > nums[m]) {
                if(nums[m] <= target && target <= nums[r])
                    l = m;
                else
                    r = m;
            }
            else
                l++;
        }
        if(nums[l] == target) return true;
        if(nums[r] == target) return true;
        return false;
    }
}