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33. Search in Rotated Sorted Array ​

题目 ​

There is an integer array nums sorted in ascending order (with distinct values).

Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].

Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.

You must write an algorithm with O(log n) runtime complexity.

 

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

Example 3:

Input: nums = [1], target = 0
Output: -1

 

Constraints:

  • 1 <= nums.length <= 5000
  • -104 <= nums[i] <= 104
  • All values of nums are unique.
  • nums is an ascending array that is possibly rotated.
  • -104 <= target <= 104
Related Topics
  • 数组
  • 二分查找

  • 👍 3194
  • 👎 0

    • 解法 ​

    主要思考有序数组的排列，获取 mid值后，可以判断 left -> mid -> right, 两个方向中，必有一个方向是有序的

    写法上采用了 [left, right) 的写法，则需要注意获取nums最右侧的值是 nums[right - 1] 而不是 nums[right]

    py
    # leetcode submit region begin(Prohibit modification and deletion)
class Solution:
    def search(self, nums: List[int], target: int) -> int:
        left = 0
        right = len(nums)

        while left < right:
            mid = (left + right) // 2

            if nums[mid] == target:
                return mid

            # 确定left -> mid是升序的
            if nums[mid] > nums[left]:
                if nums[mid] > target >= nums[left]:
                    right = mid
                else:
                    left = mid + 1
            # 确定mid -> right 是升序列的
            else:
                if nums[mid] < target <= nums[right - 1]:
                    left = mid + 1
                else:
                    right = mid

        return -1
        
# leetcode submit region end(Prohibit modification and deletion)

    复杂度分析 ​

    • 时间复杂度O(log N)
    • 空间复杂度O(1)