[Swift]LeetCode1163. 按字典序排在最后的子串 | Last Substring in Lexicographical Order

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Example 1:

Input: "abab"
Output: "bab"
Explanation: The substrings are ["a", "ab", "aba", "abab", "b", "ba", "bab"]. The lexicographically maximum substring is "bab".

Example 2:

Input: "leetcode"
Output: "tcode"

Note:

1. 1 <= s.length <= 10^5
2. s contains only lowercase English letters.

---

给你一个字符串 s，找出它的所有子串并按字典序排列，返回排在最后的那个子串。

示例 1：

输入："abab"
输出："bab"
解释：我们可以找出 7 个子串 ["a", "ab", "aba", "abab", "b", "ba", "bab"]。按字典序排在最后的子串是 "bab"。

示例 2：

输入："leetcode"
输出："tcode"

提示：

1. 1 <= s.length <= 10^5
2. s 仅含有小写英文字符。

---

 Runtime: 168 ms

Memory Usage: 22.9 MB

class Solution {
    func lastSubstring(_ s: String) -> String {
        let arrS:[Character] = Array(s)
        var i:Int = 0
        let len:Int = s.count
        for j in 1..<len
        {
            var sz:Int = 0
            while(j + sz < len)
            {
                if arrS[i + sz] == arrS[j + sz]
                {
                    sz += 1
                    continue
                }
                i = arrS[j + sz] > arrS[i + sz] ? j : i
                break                
            }
            if j + sz == len
            {
                break
            }
        }        
        return s.subString(i)
    }
}

extension String {
    // 截取字符串：从index到结束处
    // - Parameter index: 开始索引
    // - Returns: 子字符串
    func subString(_ index: Int) -> String {
        let theIndex = self.index(self.endIndex, offsetBy: index - self.count)
        return String(self[theIndex..<endIndex])
    }    
}

---

224ms 

class Solution {
    func lastSubstring(_ s: String) -> String {
                
        let chars = Array(s)
        var highest: Character = "A"
        var idxs = [Int]()
        var distinct = 0
        for i in chars.indices {
            if chars[i] > highest {
                highest = chars[i]
                idxs = [Int]()
                distinct += 1
            }

            if chars[i] == highest {
                idxs.append(i)
            } else {
                distinct += 1
            }
        }

        if distinct == 1 {
            return s
        }

        var shift = 1;

        var nextLevel = [Int]()

        while idxs.count > 1 {
            var shiftHighest: Character = "A"
            for i in idxs {
                if i + shift < chars.count {
                    if chars[i+shift] > shiftHighest {
                        shiftHighest = chars[i+shift]
                        nextLevel = [Int]()
                    }
                    if chars[i+shift] == shiftHighest {
                        nextLevel.append(i)
                    }
                }
            }
            idxs = nextLevel
            nextLevel.removeAll()
            shift += 1
        }
        return String(chars[idxs[0]..<chars.count])
    }
}

---

476ms

class Solution {
    func lastSubstring(_ s: String) -> String {
        
        var ret = s[s.startIndex...]
        
        for i in s.indices.dropFirst() {
            ret = max(ret, s[i...])
        }
        
        return String(ret)
    }
}

---

1772ms

class Solution {
    func lastSubstring(_ s: String) -> String {
        var maxCharacter = Character(Unicode.Scalar(0))
        var maxIndex = 0
        
        for (index, character) in s.enumerated() {
            if character > maxCharacter {
                maxCharacter = character
                maxIndex = index
            }
            else if character == maxCharacter, 
            String(s[String.Index(encodedOffset: index)...]) > String(s[String.Index(encodedOffset: maxIndex)...]) {
                maxCharacter = character
                maxIndex = index
            }
        }
        
        return String(s[String.Index(encodedOffset: maxIndex)...])
    }
}