LeetCode 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.

题意：给出一个数组（有重复元素，是经过旋转的有序数组）及一个数，问这个数是否在数组中出现

思路：二分法

代码如下

第一种用递归(用时364ms)

public class Solution
{private boolean check(int[] nums, int left, int right, int target){int mid = (left + right) >> 1;if (left > right) return false;if (nums[mid] == target) return true;return check(nums, left, mid - 1, target) || check(nums, mid + 1, right, target);}public boolean search(int[] nums, int target){int left = 0, right = nums.length - 1;return check(nums, left, right, target);}
}

第二种非递归（用时388ms）
代码如下

public class Solution {public boolean search(int[] nums, int target){int left = 0, right = nums.length - 1;while (left <= right){int mid = (left + right) >> 1;if (nums[mid] == target) return true;else if (nums[left] == nums[mid] && nums[mid] == nums[right]) {left++;right--;}else if (nums[left] <= nums[mid]) {if (target >= nums[left] && target < nums[mid]) right = mid - 1;else left = mid + 1;}else {if (target > nums[mid] && target <= nums[right]) left = mid + 1;else right = mid - 1;}}return false;}
}

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