LeetCode每日一题(1545. Find Kth Bit in Nth Binary String)

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本文介绍了一种生成二进制字符串序列的方法，并提供了计算第k位的高效算法。通过递归和位操作技巧，解决给定正整数n和k时，如何快速找到S_n中第k个比特位的问题。实例和约束条件展示了如何应用这些规则。

Given two positive integers n and k, the binary string Sn is formed as follows:

S1 = “0”
Si = Si - 1 + “1” + reverse(invert(Si - 1)) for i > 1
Where + denotes the concatenation operation, reverse(x) returns the reversed string x, and invert(x) inverts all the bits in x (0 changes to 1 and 1 changes to 0).

For example, the first four strings in the above sequence are:

S1 = “0”
S2 = “011”
S3 = “0111001”
S4 = “011100110110001”
Return the kth bit in Sn. It is guaranteed that k is valid for the given n.

Example 1:

Input: n = 3, k = 1
Output: “0”

Explanation: S3 is “0111001”.
The 1st bit is “0”.

Example 2:

Input: n = 4, k = 11
Output: “1”

Explanation: S4 is “011100110110001”.
The 11th bit is “1”.

Constraints:

1 <= n <= 20
1 <= k <= 2n - 1

假设 k < length / 2 则 si[k] = si-1[k], 如果 k > length / 2 则 si[k] = invert(si-1[length - k + 1])

impl Solution {
    fn rc(length: i32, k: i32, is_rev: bool) -> char {
        if length == 1 {
            if is_rev {
                return '1';
            }
            return '0';
        }
        if k == length / 2 + 1 {
            if is_rev {
                return '0';
            }
            return '1';
        }
        if k < length / 2 + 1 {
            return Solution::rc(length / 2, k, is_rev);
        }
        Solution::rc(length / 2, length - k + 1, !is_rev)
    }

    pub fn find_kth_bit(n: i32, k: i32) -> char {
        let mut length = 0;
        for _ in 0..n {
            length = length * 2 + 1;
        }
        if length == 1 {
            return '0';
        }
        if k == length / 2 + 1 {
            return '1';
        }
        if k < length / 2 + 1 {
            return Solution::rc(length / 2, k, false);
        }
        Solution::rc(length / 2, length - k + 1, true)
    }
}