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)256ms (67.37%)Ans 2 (Python) O ( N ) O(N) O(N) O ( 1 ) O(1) O(1)252ms (76.14%)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(单向遍历):
def validMountainArray(self, A: List[int]) -> bool: up = 0 down = 0 for i in range(len(A) - 1): if A[i] < A[i + 1]: if down: return False up += 1 elif A[i] > A[i + 1]: down += 1 else: return False return up > 0 and down > 0解法二(双向遍历):
def validMountainArray(self, A: List[int]) -> bool: idx1 = 0 idx2 = len(A) - 1 while idx1 < idx2: if A[idx1] < A[idx1 + 1]: idx1 += 1 elif A[idx2 - 1] > A[idx2]: idx2 -= 1 else: return False return idx1 != 0 and idx2 != len(A) - 1