Search in Rotated Sorted Array II

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?

/ 这个问题在面试中不会让实现完整程序 // 只需要举出能够最坏情况的数据是 [1,1,1,1... 1] 里有一个0即可。 // 在这种情况下是无法使用二分法的，复杂度是O(n) // 因此写个for循环最坏也是O(n)，那就写个for循环就好了 // 如果你觉得，不是每个情况都是最坏情况，你想用二分法解决不是最坏情况的情况，那你就写一个二分吧。 // 反正面试考的不是你在这个题上会不会用二分法。这个题的考点是你想不想得到最坏情况。

public boolean search(int[] A, int target) {
        for (int i = 0; i < A.length; i ++) {
            if (A[i] == target) {
                return true;
            }
        }
        return false;
    }

int start = 0, end = nums.length - 1, mid = -1;
        while(start <= end) {
            mid = (start + end) / 2;
            if (nums[mid] == target) {
                return true;
            }
            //If we know for sure right side is sorted or left side is unsorted
            if (nums[mid] < nums[end] || nums[mid] < nums[start]) {
                if (target > nums[mid] && target <= nums[end]) {
                    start = mid + 1;
                } else {
                    end = mid - 1;
                }
            //If we know for sure left side is sorted or right side is unsorted
            } else if (nums[mid] > nums[start] || nums[mid] > nums[end]) {
                if (target < nums[mid] && target >= nums[start]) {
                    end = mid - 1;
                } else {
                    start = mid + 1;
                }
            //If we get here, that means nums[start] == nums[mid] == nums[end], then shifting out
            //any of the two sides won't change the result but can help remove duplicate from
            //consideration, here we just use end-- but left++ works too
            } else {
                end--;
            }
        }
        
        return false;