[LeetCode] 747. Largest Number At Least Twice of Others

这道题思路很简单，就是在一个loop下找最大的和第二大的数值。记住，python最小的数值是-sys.maxsize-1

class Solution:
    def dominantIndex(self, nums: List[int]) -> int:
        if not nums or len(nums) == 1:
            return 0

        largest, sec_largest, idx = -sys.maxsize - 1, -sys.maxsize - 1, 0

        for i in range(len(nums)):
            if nums[i] >= largest:
                sec_largest = largest
                largest = nums[i]
                idx = i
            if nums[i] > sec_largest and nums[i] < largest:
                sec_largest = nums[i]

        if largest >= 2 * sec_largest:
            return idx
        return -1
            

原文地址：https://www.cnblogs.com/codingEskimo/p/12154613.html