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 ( N ) O(N) O(N)96ms (79.94%)Ans 2 (Python) O ( N ) O(N) O(N) O ( N ) O(N) O(N)100ms (63.23%)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(Pythonic):
def sortArrayByParity(self, A: List[int]) -> List[int]: return [a for a in A if a % 2 == 0] + [a for a in A if a % 2 == 1]解法二:
def sortArrayByParity(self, A: List[int]) -> List[int]: odd = [] even = [] for a in A: if a % 2 == 0: even.append(a) else: odd.append(a) return even+odd