LeetCode题解(0696)：二进制字符串中0和1数量相同的连续子串数量(Python)

题目：原题链接（简单）

标签：字符串

解法时间复杂度空间复杂度执行用时Ans 1 (Python)$O(N)$$O(1)$252ms (42.00%)Ans 2 (Python)$O(N)$$O(1)$104ms (100.00%)Ans 3 (Python)

解法一：

def countBinarySubstrings(self, s: str) -> int: num0 = 0 num1 = 0 ans = 0 for i in range(len(s)): n = s[i] if i > 0 and s[i] != s[i - 1]: if s[i] == "0": num0 = 0 else: num1 = 0 if n == "1": if num0 > 0: ans += 1 num0 -= 1 num1 += 1 elif n == "0": if num1 > 0: ans += 1 num1 -= 1 num0 += 1 return ans

解法二（减少判断次数）：

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def countBinarySubstrings(self, s: str) -> int: pre = 0 count = 1 curr = s[0] ans = 0 for n in s[1:]: if n == curr: count += 1 else: pre = count count = 1 curr = n if pre >= count: ans += 1 return ans