题目:原题链接(简单)
解法时间复杂度空间复杂度执行用时Ans 1 (Python) O ( N ) O(N) O(N) O ( 1 ) O(1) O(1)44ms (86.21%)Ans 2 (Python) O ( N ) O(N) O(N) O ( 1 ) O(1) O(1)44ms (86.21%)Ans 3 (Python)LeetCode的Python执行用时随缘,只要时间复杂度没有明显差异,执行用时一般都在同一个量级,仅作参考意义。
解法一:
def dominantIndex(self, nums: List[int]) -> int: if len(nums) == 1: return 0 max1 = -1 max2 = -1 idx = -1 for i in range(len(nums)): n = nums[i] if n > max1: max2 = max1 max1 = n idx = i elif n > max2: max2 = n if max1 >= max2 * 2: return idx else: return -1解法二:
def dominantIndex(self, nums: List[int]) -> int: m = max(nums) if all([m >= 2 * n for n in nums if n != m]): return nums.index(m) else: return -1