Why not solve the problem directly by comparing each element with the first?


  • 0
    M
    class Solution(object):
    def findMin(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if len(nums)==0: return
        current_min = nums[0]
        for num in nums:
            if num<current_min:
                return num
        return current_min
    

    The answer is simply the first element or the first following element less than the first one. It seems much easier than the solution.


  • 1
    C

    Because that would be O(N) worse time complexity. The optimal solution uses binary search and achieves O(log N) time this way.


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