[Swift]LeetCode796. 旋转字符串 | Rotate String

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We are given two strings, A and B.

A shift on A consists of taking string A and moving the leftmost character to the rightmost position. For example, if A = 'abcde', then it will be 'bcdea' after one shift on A. Return True if and only if A can become B after some number of shifts on A.

Example 1:
Input: A = 'abcde', B = 'cdeab'
Output: true

Example 2:
Input: A = 'abcde', B = 'abced'
Output: false

Note:

• A and B will have length at most 100.

---

给定两个字符串, A 和 B。

A 的旋转操作就是将 A 最左边的字符移动到最右边。 例如, 若 A = 'abcde'，在移动一次之后结果就是'bcdea' 。如果在若干次旋转操作之后，A 能变成B，那么返回True。

示例 1:
输入: A = 'abcde', B = 'cdeab'
输出: true

示例 2:
输入: A = 'abcde', B = 'abced'
输出: false

注意：

• A 和 B 长度不超过 100。

---

Runtime: 4 ms

Memory Usage: 20.2 MB

class Solution {
    func rotateString(_ A: String, _ B: String) -> Bool {
        if A.isEmpty && B.isEmpty {return true}
        if A.isEmpty && !B.isEmpty {return false}
        if !A.isEmpty && B.isEmpty {return false}
        return A.count == B.count && (A + A).contains(B)
    }
}

---

4ms

class Solution {
    func rotateString(_ A: String, _ B: String) -> Bool {
        guard A.count == B.count else { return false }
        guard !A.isEmpty && !B.isEmpty else { return true }
        return (A + A).contains(B)
    }
}

---

8ms

class Solution {
    func rotateString(_ A: String, _ B: String) -> Bool {
        
        if A.count == 0 && B.count == 0 {
            return true
        }

        var A = A

        for _ in 0..<A.count {

            if A == B {
                return true
            }

            let index = A.index(A.startIndex, offsetBy: 0)
            A.append(A[index])
            A.removeFirst()
        }

        return false
    }
}

---

16ms

class Solution {
    func rotateString(_ A: String, _ B: String) -> Bool {
        
        guard A.count == B.count else {
            return false
        }
        
        var A = A
        
        for _ in 0..<A.count where A != B {
            A.append(A.removeFirst())
        }
        
        return A == B
    }
}

---

20016kb

class Solution {
    func rotateString(_ A: String, _ B: String) -> Bool {
        guard A.length == B.length else { return false }
        guard A != B else { return true }
        guard B.length > 0 else { return false }
        guard A.length > 0 else { return false }
        
        let chars = Array(A).map({ String($0) })
        let n = chars.count
        var fullRotation = [String](repeating:" ", count: 2 * n - 1)
        
        for i in 0..<n {
            fullRotation[i + n - 1] = chars[i]
        }
        for i in (1..<n).reversed() {
            fullRotation[i - 1] = chars[i]
        }
        
        return fullRotation.joined().contains(B)
    }
}