LeetCode 81 Search in Rotated Sorted Array II

最新推荐文章于 2025-08-22 19:21:44 发布

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本文探讨了在允许重复元素的旋转排序数组中搜索特定目标值的问题，并提供了一种高效的搜索算法实现。该算法通过判断数组的有序部分来决定搜索方向，有效避免了重复元素带来的搜索复杂性增加。

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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.

	public boolean search2(int[] nums, int target) {
		int low = 0, high = nums.length - 1, mid;
		while (low <= high) {
			mid = low + (high - low) / 2;
			if (nums[mid] == target) return true;
			while (nums[mid] == nums[low] && mid > low) low++;
			while (nums[mid] == nums[high] && high > mid) high--;
			if (nums[mid] <= nums[high]) {//右边有序
				if (nums[mid] < target && target <= nums[high]) low = mid + 1;
				else high = mid - 1;
			} else {//左边有序
				if (nums[low] <= target && target < nums[mid]) high = mid - 1;
				else low = mid + 1;
			}
		}
		return false;
	}