8/15 每日一題(二分法)

 Given an array of integers nums which is sorted in ascending order, and an integer target, write a function to search target in nums. If target exists, then return its index. Otherwise, return -1.

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

 

Example 1:

Input: nums = [-1,0,3,5,9,12], target = 9
Output: 4
Explanation: 9 exists in nums and its index is 4

Example 2:

Input: nums = [-1,0,3,5,9,12], target = 2
Output: -1
Explanation: 2 does not exist in nums so return -1

 

Constraints:

• 1 <= nums.length <= 104
• -104 < nums[i], target < 104
• All the integers in nums are unique.
• nums is sorted in ascending order.

class Solution:

    def search(self, nums: List[int], target: int) -> int:

        left_n=0

        right_n=len(nums)-1

        mid_n=(left_n+right_n)//2

       

        while left_n <= right_n:

            if nums[mid_n] == target:

                return mid_n

           

            elif nums[mid_n] < target:

                left_n=mid_n+1

               

            else:

                right_n=mid_n-1

            mid_n=(left_n+right_n)//2

        return -1