What is the solution approach for 'Check if Binary String Has at Most One Segment of Ones'?

The solution involves scanning the string to identify contiguous segments of '1's and counting them. If more than one segment is found, return false.

How does the string-driven approach help in solving this problem?

The string-driven approach allows you to focus on simple, efficient traversal of the binary string, making it easy to identify and count segments of '1's.

Are there any edge cases I should consider for this problem?

Yes, consider cases where the string consists of all '0's, all '1's, or just a single character. These cases may behave differently in your approach.

What is the time complexity of the solution?

The time complexity is O(n), where n is the length of the string, as we scan the string once to count the segments of '1's.

Can this problem be solved with a different approach?

While the string-driven approach is optimal, alternative methods might involve more complex data structures or a recursive approach. However, these might not be as efficient.

#1784

Easy

auto_awesomeString-driven solution strategy

LeetCode 题解工作台

检查二进制字符串字段

给你一个二进制字符串 s ，该字符串不含前导零。如果 s 包含零个或一个由连续的 '1' 组成的字段，返回 true 。否则，返回 false 。示例 1：输入： s = "1001" 输出： false 解释：由连续若干个 '1' 组成的字段数量为 2，返回 false 示…

字符串

题目描述

给你一个二进制字符串 s ，该字符串 不含前导零 。

如果 s 包含 零个或一个由连续的 '1' 组成的字段 ，返回 true 。否则，返回 false 。

示例 1：

输入：s = "1001"
输出：false
解释：由连续若干个 '1' 组成的字段数量为 2，返回 false

示例 2：

输入：s = "110"
输出：true

提示：

1 <= s.length <= 100
s[i] 为 '0' 或 '1'
s[0] 为 '1'

lightbulb

解题思路

方法一：脑筋急转弯

注意到字符串 $s$ 不含前导零，说明 $s$ 以 '1' 开头。

若字符串 $s$ 存在 "01" 串，那么 $s$ 就是形如 "1...01..." 的字符串，此时 $s$ 出现了至少两个连续的 '1' 片段，不满足题意，返回 $\textit{false}$ 。

若字符串 $s$ 不存在 "01" 串，那么 $s$ 只能是形如 "1..1000..." 的字符串，此时 $s$ 只有一个连续的 '1' 片段，满足题意，返回 $\textit{true}$ 。

因此，只需要判断字符串 $s$ 是否存在 "01" 串即可。

时间复杂度 $O(n)$ 。其中 $n$ 为字符串 $s$ 的长度。空间复杂度 $O(1)$ 。

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class Solution:
    def checkOnesSegment(self, s: str) -> bool:
        return '01' not in s

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Focus on counting segments of '1's efficiently.
question_mark
Ensure the approach handles edge cases like all '1's or '0's.
question_mark
Evaluate whether a more optimized solution can reduce redundant checks.

warning

常见陷阱

外企场景

error
Miscounting the segments due to improper string scanning.
error
Not handling edge cases like strings with no '1's or strings where all characters are '1'.
error
Overcomplicating the solution by introducing unnecessary data structures.

swap_horiz

进阶变体

外企场景

arrow_right_alt
What if the string has leading or trailing zeros? The logic can be adapted to handle these cases effectively.
arrow_right_alt
How would the solution change if we allowed the string to contain other characters? This introduces complexity in the segmentation logic.
arrow_right_alt
Can we optimize further for strings with very large lengths? Explore constant space solutions.

help

常见问题

外企场景

继续练习

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