{"id":8209,"date":"2021-03-06T22:24:55","date_gmt":"2021-03-07T06:24:55","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=8209"},"modified":"2021-03-06T22:25:17","modified_gmt":"2021-03-07T06:25:17","slug":"leetcode-1784-check-if-binary-string-has-at-most-one-segment-of-ones","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/string\/leetcode-1784-check-if-binary-string-has-at-most-one-segment-of-ones\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1784. Check if Binary String Has at Most One Segment of Ones"},"content":{"rendered":"\n

Given a binary string s<\/code> \u200b\u200b\u200b\u200b\u200bwithout leading zeros<\/strong>, return true<\/code>\u200b\u200b\u200b if <\/em>s<\/code> contains at most one contiguous segment of ones<\/strong><\/em>. Otherwise, return false<\/code>.<\/p>\n\n\n\n

Example 1:<\/strong><\/p>\n\n\n\n

Input:<\/strong> s = \"1001\"\nOutput:<\/strong> false\nExplanation: <\/strong>The ones do not form a contiguous segment.\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> s = \"110\"\nOutput:<\/strong> true<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • 1 <= s.length <= 100<\/code><\/li>
  • s[i]<\/code>\u200b\u200b\u200b\u200b is either '0'<\/code> or '1'<\/code>.<\/li>
  • s[0]<\/code> is '1'<\/code>.<\/li><\/ul>\n\n\n\n
    Solution: Counting<\/strong><\/h2>\n\n\n\n
    increase counter if s[i] == ‘1’ otherwise, reset counter.
    increase counts when counter becomes 1.
    return counts == 1<\/p>\n\n\n\n
    Time complexity: O(n)
    Space complexity: O(1)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \/\/ Author: Huahua\nclass Solution {\npublic:\n  bool checkOnesSegment(string s) {\n    int count = 0;\n    int ones = 0;\n    for (const char c : s) {\n      if (c == '1') {\n        count += (++ones == 1);        \n      } else {\n        ones = 0;\n      }\n    }\n    return count == 1;\n  }\n};\n<\/pre>\n<\/div><\/div>\n\n\n\n
    <\/p>\n","protected":false},"excerpt":{"rendered":"
    Given a binary string s \u200b\u200b\u200b\u200b\u200bwithout leading zeros, return true\u200b\u200b\u200b if s contains at most one contiguous segment of ones. Otherwise, return false. Example 1: Input: s = “1001” Output: false Explanation: The…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[47],"tags":[4],"class_list":["post-8209","post","type-post","status-publish","format-standard","hentry","category-string","tag-string","entry","simple"],"_links":{"self":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8209","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/comments?post=8209"}],"version-history":[{"count":2,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8209\/revisions"}],"predecessor-version":[{"id":8211,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/posts\/8209\/revisions\/8211"}],"wp:attachment":[{"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/media?parent=8209"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/categories?post=8209"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/zxi.mytechroad.com\/blog\/wp-json\/wp\/v2\/tags?post=8209"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}