LeetCode每日一题(2380. Time Needed to Rearrange a Binary String)

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这篇博客探讨了一个关于二进制字符串的问题，其中每次操作将所有出现的 '01' 对同时替换为 '10'。文章通过示例解释了这个过程，并提供了一个解决方案来计算完成这一过程所需的最少秒数。例如，对于输入字符串 '0110101'，在4秒内可以完成转换。博客内容涉及动态规划和字符串处理算法，适合对算法和二进制处理感兴趣的读者。

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.

Example 1:

Input: s = “0110101”
Output: 4

Explanation:
After one second, s becomes “1011010”.
After another second, s becomes “1101100”.
After the third second, s becomes “1110100”.
After the fourth second, s becomes “1111000”.
No occurrence of “01” exists any longer, and the process needed 4 seconds to complete,
so we return 4.

Example 2:

Input: s = “11100”
Output: 0

Explanation:
No occurrence of “01” exists in s, and the processes needed 0 seconds to complete,
so we return 0.

Constraints:

1 <= s.length <= 1000
s[i] is either ‘0’ or ‘1’.

dp[i]为整理 s[0…=i]所需要的秒数, zeros 为到 i 为止的 0 的数量，如果 s[i+1] == ‘1’, 则 dp[i+1] = max(zeros, dp[i] + 1), 如果 s[i+1] == ‘0’, 则 dp[i+1] = dp[i-1]。


impl Solution {
    pub fn seconds_to_remove_occurrences(s: String) -> i32 {
        let chars: Vec<char> = s.chars().collect();
        let mut dp = vec![0; s.len()];
        let mut zeros = if chars[0] == '1' { 0 } else { 1 };
        for i in 1..s.len() {
            if chars[i] == '0' {
                zeros += 1;
                dp[i] = dp[i - 1];
                continue;
            }
            if zeros > 0 {
                dp[i] = (dp[i - 1] + 1).max(zeros);
                continue;
            }
            dp[i] = dp[i - 1];
        }
        *dp.last().unwrap()
    }
}