LeetCode #747. Largest Number At Least Twice of Others

题目

747. Largest Number At Least Twice of Others

解题方法

遍历数组，维护最大值maxnum、次大值的两倍twice、最大值位置maxnumpos三个变量：若当前位置大于maxnum，则令twice为maxnum的两倍，更新maxnum和maxnumpos；否则，令twice = max(twice, nums[i] * 2)，循环结束后判断maxnum和twice的关系即可。
时间复杂度：O(n)
空间复杂度：O(1)

代码

class Solution:
    def dominantIndex(self, nums: List[int]) -> int:
        maxnum = nums[0]
        twice = 0
        maxnumpos = 0
        for i in range(1, len(nums)):
            if nums[i] > maxnum:
                twice = maxnum * 2
                maxnum = nums[i]
                maxnumpos = i
            else:
                twice = max(twice, nums[i] * 2)
        if maxnum >= twice:
            return maxnumpos
        else:
            return -1

标签：maxnum,nums,复杂度,747,Number,twice,Others,maxnumpos
来源： https://www.cnblogs.com/RatsCommander/p/14009114.html