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)––32ms (94.29%)Ans 2 (Python) O ( 1 ) O(1) O(1) O ( 1 ) O(1) O(1)36ms (84.38%)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 hasAlternatingBits(self, n: int) -> bool: n = bin(n) return "11" not in n and "00" not in n解法二(位运算):
def hasAlternatingBits(self, n: int) -> bool: n = n ^ (n << 1) # -> 1...10 or 1...11 n = n >> 1 # -> 1..11 return n & (n + 1) == 0