[LeetCode] 704. Binary Search_Easy tag: Binary Search

Given a sorted (in ascending order) integer array nums of n elements and a target value, write a function to search target in nums. If target exists, then return its index, otherwise return -1.

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

 

Note:

1. You may assume that all elements in nums are unique.
2. n will be in the range [1, 10000].
3. The value of each element in nums will be in the range [-9999, 9999].

 

因为是常规的没有duplicates的binary search, 所以参考[LeetCode] questions conclusion_ Binary Search1.1的思路和code即可.

Code

class Solution:
    def search(self, nums, target):
        l, r = 0, len(nums)-1
        if target < nums[0] or target > nums[-1]: return -1
        while l + 1 < r:
            mid = l + (r-l)//2   # l + r maybe overflow, above 2**32
            if nums[mid] > target:
                r = mid
            elif nums[mid] < target:
                l = mid
            else:
                return mid
        
        if nums[l] == target:
            return l
        if nums[r] == target:
            return r
        return -1