[Swift]LeetCode14. 最长公共前缀 | Longest Common Prefix

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Write a function to find the longest common prefix string amongst an array of strings.

If there is no common prefix, return an empty string "".

Example 1:

Input: ["flower","flow","flight"]
Output: "fl"

Example 2:

Input: ["dog","racecar","car"]
Output: ""
Explanation: There is no common prefix among the input strings.

Note:

All given inputs are in lowercase letters a-z.


编写一个函数来查找字符串数组中的最长公共前缀。

如果不存在公共前缀,返回空字符串 ""

示例 1:

输入: ["flower","flow","flight"]
输出: "fl"

示例 2:

输入: ["dog","racecar","car"]
输出: ""
解释: 输入不存在公共前缀。

说明:

所有输入只包含小写字母 a-z 。


16ms

 1 class Solution {
 2     func longestCommonPrefix(_ strs: [String]) -> String {
 3         let minStr = try? strs.min(by: { $0.count < $1.count })
 4         guard var minS = minStr! else {
 5             return ""
 6         }
 7         
 8         var endIndex = minS.endIndex
 9         for str in strs {
10             if (str.hasPrefix(minS)) {
11                 continue
12             }
13             while endIndex > minS.startIndex && !str.hasPrefix(minS) {
14                 endIndex = minS.index(before: endIndex)
15                 minS = String(minS[minS.startIndex..<endIndex])
16             }
17         }
18         return minS
19     }
20 }

20ms

 1 class Solution {
 2        func longestCommonPrefix(_ strs: [String]) -> String {
 3         
 4         if strs.count == 0 {
 5             return ""
 6         }
 7         var prefix = strs[0]
 8         for i in 0..<strs.count {
 9             while strs[i].hasPrefix(prefix) == false {
10                 prefix = prefix.substring(to: prefix.index(before: prefix.endIndex))
11                 if prefix.isEmpty {
12                     return ""
13                 }
14             }
15         }
16         return prefix
17     }
18 }

24ms

 1 class Solution {
 2     func longestCommonPrefix(_ strs: [String]) -> String {
 3         guard strs.count > 0 else { return "" }
 4         var shortestPrefix = strs.min{ $0.count < $1.count }!
 5         print(shortestPrefix)
 6         for s in strs {
 7             while shortestPrefix.count > 0 , !s.hasPrefix(shortestPrefix) {
 8                 shortestPrefix.removeLast()
 9             }
10         }
11         return shortestPrefix
12     }
13 }

28ms

 1 class Solution {
 2     func longestCommonPrefix(_ strs: [String]) -> String {
 3 
 4         var res = ""
 5         if strs.isEmpty {
 6             
 7             return res
 8         }
 9         
10         res = strs[0]
11         
12         for i in 1..<strs.count {
13             
14             var find = Array(strs[i])
15             var current = Array(res)
16             res = ""
17             for i in 0..<find.count {
18                 
19                 if i >= current.count {
20                     break
21                 }
22                 
23                 if current[i] != find[i] {
24                     
25                     break
26                 } else {
27                     
28                     res += String(current[i])
29                 }
30             }
31             
32         }
33         
34         return res
35     }
36 }

32ms

 1 class Solution {
 2   func longestCommonPrefix(_ strs: [String]) -> String {
 3     guard let first = strs.first else { return "" }
 4 
 5     var currentPrefix = ""
 6     var bestPrefix = ""
 7 
 8     for letter in first { 
 9       currentPrefix.append(letter)
10 
11       for word in strs {
12         if !word.hasPrefix(currentPrefix) { 
13           return bestPrefix 
14         }
15       }
16 
17       bestPrefix = currentPrefix
18     }
19 
20     return bestPrefix                
21   }
22 }
原文地址:https://www.cnblogs.com/strengthen/p/9697903.html