# LeetCode 1784. Check if Binary String Has at Most One Segment of Ones [LeetCode 1784. Check if Binary String Has at Most One Segment of Ones](https://leetcode.com/problems/check-if-binary-string-has-at-most-one-segment-of-ones/) (<font color=#00AF9B>Easy</font> 39.2%) <!-- (<font color=#00AF9B>Easy</font> 53.8%) (<font color=#FFB800>Medium</font> 39.6%) (<font color=#FF375F>Hard</font>) --> - 限制 : <ul> <li><code>1 <= s.length <= 100</code></li> <li><code>s[i] is either '0' or '1'.</code></li> <li><code>s[0] is '1'.</code></li> </ul> - Solution 這題要找的是1有沒有都在字串的前面,如果沒有就回傳 false。 - 時間複雜度: $O(n)$ - 空間複雜度: $O(1)$ - 程式碼 ```c++= class Solution { public: bool checkOnesSegment(string s) { char nowChr = '1'; for (auto& chr : s) { if (chr == '0') { nowChr = chr; continue; } if (chr != nowChr) return false; } return true; } }; ``` </details>