You are given a binary string s. In one second, all occurrences of "01" are simultaneously replaced with "10". This process repeats until no occurrences of "01" exist.
Return the number of seconds needed to complete this process.
\n\n\n
Example 1:
\n\n\nInput: s = "0110101"\nOutput: 4\nExplanation: \nAfter one second, s becomes "1011010".\nAfter another second, s becomes "1101100".\nAfter the third second, s becomes "1110100".\nAfter the fourth second, s becomes "1111000".\nNo occurrence of "01" exists any longer, and the process needed 4 seconds to complete,\nso we return 4.\n\n\n
Example 2:
\n\n\nInput: s = "11100"\nOutput: 0\nExplanation:\nNo occurrence of "01" exists in s, and the processes needed 0 seconds to complete,\nso we return 0.\n\n\n
\n
Constraints:
\n\n1 <= s.length <= 1000s[i] is either '0' or '1'.\n
Follow up:
\n\nCan you solve this problem in O(n) time complexity?
\n", "translatedTitle": "二进制字符串重新安排顺序需要的时间", "translatedContent": "给你一个二进制字符串 s 。在一秒之中，所有 子字符串 \"01\" 同时 被替换成 \"10\" 。这个过程持续进行到没有 \"01\" 存在。
请你返回完成这个过程所需要的秒数。
\n\n\n\n
示例 1：
\n\n\n输入：s = \"0110101\"\n输出：4\n解释：\n一秒后，s 变成 \"1011010\" 。\n再过 1 秒后，s 变成 \"1101100\" 。\n第三秒过后，s 变成 \"1110100\" 。\n第四秒后，s 变成 \"1111000\" 。\n此时没有 \"01\" 存在，整个过程花费 4 秒。\n所以我们返回 4 。\n\n\n
示例 2：
\n\n\n输入：s = \"11100\"\n输出：0\n解释：\ns 中没有 \"01\" 存在，整个过程花费 0 秒。\n所以我们返回 0 。\n\n\n
\n\n
提示：
\n\n1 <= s.length <= 1000s[i] 要么是 '0' ，要么是 '1' 。\n\n
进阶：
\n\n你能以 O(n) 的时间复杂度解决这个问题吗？
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