LeetCode 81. Search in Rotated Sorted Array II

原题链接在这里：https://leetcode.com/problems/search-in-rotated-sorted-array-ii/

题目：

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.

题解：

是Search in Rotated Sorted Array的变形。与Find Minimum in Rotated Sorted Array 和 Find Minimum in Rotated Sorted Array II的关系十分相似。

加上了duplicates, 使得原有的通过nums[mid] 和 边界值 如nums[r]的大小关系判断前后两部分那部分带有rotation的方法不再有效。e.g. 若是nums[mid] == nums[r]此时不知道前后两部哪一部分有rotation. 所以此时只能移动边界, 如 r--. 所以worst case 是全部相同, 此时Time Complexity O(n).

其他情况若是 nums[mid] 和 nums[r] 不相同，就能判断出前后哪一部分有rotation, 剩下的与Search in Rotated Sorted Array相同。

Time Complexity: O(n). Space: O(1).

AC Java:

class Solution {
    public boolean search(int[] nums, int target) {
        if(nums == null || nums.length == 0){
            return false;
        }

        int l = 0;
        int r = nums.length - 1;
        while(l <= r){
            int mid = l + (r - l) / 2;
            if(nums[mid] == target){
                return true;
            }
            
            if(nums[mid] < nums[r]){
                if(nums[mid] <= target && target <= nums[r]){
                    l = mid + 1;
                }else{
                    r = mid - 1;
                }
            }else if(nums[mid] > nums[r]){
                if(nums[l] <= target && target <= nums[mid]){
                    r = mid - 1;
                }else{
                    l = mid + 1;
                }
            }else{
                r--;
            }
        }

        return false;
    }
}