[Swift]LeetCode459. 重复的子字符串 | Repeated Substring Pattern

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Given a non-empty string check if it can be constructed by taking a substring of it and appending multiple copies of the substring together. You may assume the given string consists of lowercase English letters only and its length will not exceed 10000.、 

Example 1:

Input: "abab"
Output: True
Explanation: It's the substring "ab" twice.

Example 2:

Input: "aba"
Output: False

Example 3:

Input: "abcabcabcabc"
Output: True
Explanation: It's the substring "abc" four times. (And the substring "abcabc" twice.)

---

给定一个非空的字符串，判断它是否可以由它的一个子串重复多次构成。给定的字符串只含有小写英文字母，并且长度不超过10000。

示例 1:

输入: "abab"

输出: True

解释: 可由子字符串 "ab" 重复两次构成。

示例 2:

输入: "aba"

输出: False

示例 3:

输入: "abcabcabcabc"

输出: True

解释: 可由子字符串 "abc" 重复四次构成。 (或者子字符串 "abcabc" 重复两次构成。)

---

148ms

class Solution {
    //kmp算法
    func repeatedSubstringPattern(_ s: String) -> Bool {
        var arr:[Character] = [Character]()
        for char in s.characters
        {
            arr.append(char)
        }
        var i:Int = 1
        var j:Int = 0
        var n:Int = s.count
        var dp:[Int] = [Int](repeating:0,count:n + 1)
        while(i < n)
        {
            if arr[i] == arr[j]
            {
                i += 1
                j += 1
                dp[i] = j
            }
            else if j == 0
            {
                i += 1
            }
            else
            {
                j = dp[j]
            }
        }
        return dp[n] % (n - dp[n]) == 0 && dp[n] != 0
    }
}

---

292ms

class Solution {
    func repeatedSubstringPattern(_ s: String) -> Bool {
        let ss = s + s
        let str = ss[ss.index(after: ss.startIndex)..<ss.index(before: ss.endIndex)]
        return str.contains(s)
    }
}

---

480ms

class Solution {
    func repeatedSubstringPattern(_ s: String) -> Bool {
        let length = s.count
        var index  = length / 2

        while index >= 1 {
            if length % index == 0 {
                let c = length / index
                var current = ""
                
                for _ in 0..<c {
                    
                    let offset = s.index(s.startIndex, offsetBy: index)
                    current += String(s[..<offset])

                }
                if current == s {
                    return true
                }

            }
            index -= 1
        }
 
        return false
    }
}

---

500ms

class Solution {
    func repeatedSubstringPattern(_ s: String) -> Bool {
        let chas = [Character](s)
        let res = String(chas[1...]) + String(chas[..<(chas.count-1)])
        
        return res.contains(s)
    }
}

---

604ms

class Solution {
    func repeatedSubstringPattern(_ s: String) -> Bool {
        let count = s.count
        var huff = count / 2
        while huff >= 1 {
            if count % huff == 0 {
                let toIndex = s.index(s.startIndex, offsetBy: huff)
                let subString = s[s.startIndex..<toIndex]
                
                var num = count / huff
                var sumString = ""
                
                while num > 0 {
                    sumString = sumString + subString
                    num = num - 1
                }
                
                if sumString == s {
                    return true
                }
            }
            
            huff = huff - 1
        }
        return false
    }
}

---

3292ms

class Solution {
    func repeatedSubstringPattern(_ s: String) -> Bool {
        let length = s.count
        
        var result = false;
        for index in 1...length {
            // 整除则对比
            if length % (index) == 0 {
                // 从0到index
                let character = s.prefix(index)
                let increment = index;
                var start = increment;
                
                var isEqual = false;
                while (start < length) {
                    let begin = s.index(s.startIndex, offsetBy: start)
                    let stop = s.index(s.startIndex, offsetBy: start + increment)
                    let temp = s[begin..<stop]
                    
                    if (character == temp) {
                        start += increment;
                        isEqual = true;
                        continue;
                    } else {
                        isEqual = false;
                        break;
                    }
                }
                result = isEqual;
                if isEqual {
                    break;
                }
            }
        }
        return result
    }
}