LeetCode每日一题(2311. Longest Binary Subsequence Less Than or Equal to K)

You are given a binary string s and a positive integer k.

Return the length of the longest subsequence of s that makes up a binary number less than or equal to k.

Note:

The subsequence can contain leading zeroes.
The empty string is considered to be equal to 0.
A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.

Example 1:

Input: s = “1001010”, k = 5
Output: 5

Explanation: The longest subsequence of s that makes up a binary number less than or equal to 5 is “00010”, as this number is equal to 2 in decimal.
Note that “00100” and “00101” are also possible, which are equal to 4 and 5 in decimal, respectively.
The length of this subsequence is 5, so 5 is returned.

Example 2:

Input: s = “00101001”, k = 1
Output: 6

Explanation: “000001” is the longest subsequence of s that makes up a binary number less than or equal to 1, as this number is equal to 1 in decimal.
The length of this subsequence is 6, so 6 is returned.

Constraints:

• 1 <= s.length <= 1000
• s[i] is either ‘0’ or ‘1’.
• 1 <= k <= 109

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做到最后发现这是一道脑筋急转弯， 先说思路， 二进制从右往左每个位上的 1 分别代表 2 的 0 次方、2 的 1 次方、2 的 2 次方。。。。。， 我们每次从 s 的末尾拿一个字符，如果是 1 的话，如果我们要放到我们的结果字符串当中， 就相当于给当前值加了 2 的 n 次方， 这个 n 是当前结果字符串的长度， 如果不放当然值也不增加，长度也不增加。而如果拿出的字符是 0 的话，我们可以放心大胆的把它放到结果字符串中，因为它不会使当前结果字符串的值增加，只会增加长度。所以整体来看， 我们遇到 1 的时候应该是能增就增，因为此时如果你不增，后面结果变长了你更增不起，而如果遇到 0 则只需直接放到结果里就好，因为当前位你不是放 0 就是放 1, 你如果放弃眼前的 0 改等后面的 1，届时同样是增长 1 位，但是放 1 的话值也会相应增加的。

再说脑筋急转弯的问题， 题目里 k 的类型是 i32，也就是最多有 32 个 1, 但是题目里给出的 s.length <= 1000, 所以在我们攒齐了 32 个 1 之后，我们就不需要再去检查 1 的情况了， 只把后面所有的 0 收集起来就好了

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impl Solution {
    pub fn longest_subsequence(mut s: String, k: i32) -> i32 {
        let mut len = 0i64;
        let mut val = 0i64;
        while let Some(c) = s.pop() {
            if c == '1' {
                if len > 32 {
                    continue;
                }
                let v = 2i64.pow(len as u32);
                if val + v <= k as i64 {
                    val += v;
                    len += 1;
                }
                continue;
            }
            len += 1;
        }
        len as i32
    }
}